Part 2:
To determine an equation that is perpendicular to the line equation y = -3/2x + 12, get the negative reciprocal of the slope of the line equation.
[tex]\begin{gathered} \text{Given slope: }m=-\frac{3}{2} \\ \\ \text{The negative reciprocal is} \\ -\Big(-\frac{3}{2}\Big)^{-1}=\frac{2}{3} \\ \\ \text{We can now assume that any line in the form} \\ y=\frac{2}{3}x+b \\ \text{where }b\text{ is the y-intercept} \\ \text{is perpendicular to the line }y=-\frac{3}{2}x+12 \end{gathered}[/tex]Part 3:
An equation that is parallel to the line y = -3/2x + 12, is a line equation that will have the same slope as the original line.
Given that the slope of the line is m = -3/2, then any line equation in the form
[tex]\begin{gathered} y=-\frac{3}{2}x+b \\ \text{where} \\ b\text{ is the y-intercept} \end{gathered}[/tex]The points (1,7) and (7,5) fall on a particular line. What is its equation in point-slope form?
Use one of the specified points in your equation. Write your answer using integers, proper fractions, and improper fractions. Simplify all fractions.
Answer:
[tex]y-7=-\dfrac{1}{3}(x-1)[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{4.4cm}\underline{Slope Formula}\\\\Slope $(m)=\dfrac{y_2-y_1}{x_2-x_1}$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ \\are two points on the line.\\\end{minipage}}[/tex]
To find the equation of a line that passes through two given points, first find its slope by substituting the given points into the slope formula.
Define the points:
(x₁, y₁) = (1, 7)(x₂, y₂) = (7, 5)Substitute the points into the slope formula:
[tex]\implies m=\dfrac{5-7}{7-1}=\dfrac{-2}{6}=-\dfrac{1}{3}[/tex]
Therefore, the slope of the line is -¹/₃.
[tex]\boxed{\begin{minipage}{5.8 cm}\underline{Point-slope form of a linear equation}\\\\$y-y_1=m(x-x_1)$\\\\where:\\ \phantom{ww}$\bullet$ $m$ is the slope. \\ \phantom{ww}$\bullet$ $(x_1,y_1)$ is a point on the line.\\\end{minipage}}[/tex]
To find the equation in point-slope form, simply substitute the found slope and one of the given points into the point-slope formula:
[tex]\implies y-7=-\dfrac{1}{3}(x-1)[/tex]
How to find the area of a regular hexagon with a radius of 12 inches? Please help
Slope of Linear EquationsWhich description best compares the graph given by the following equations:23-5y = 82Y == -6Choose one. 4 pointsO parallelO perpendicularintersecting but not perpendicularO coinciding
Answer:
The two lines are parallel.
Explanation:
We have the equations:
[tex]\begin{gathered} 2x-5y=8 \\ y=\frac{2}{5}x-6 \end{gathered}[/tex]Let's solve the first one for y, so we get the same formatting on both euqations:
[tex]\begin{gathered} 2x-5y=8 \\ 5y=2x-8 \\ y=\frac{2}{5}x-\frac{8}{5} \end{gathered}[/tex]SInce the two lines have the same slope, 2/5, the two lines are parallel.
(Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) And determine the quadrants of A+B and A-B.
Given that:
[tex]\cos A=\frac{5}{13}[/tex]Where:
[tex]0And:[tex]\cos B=\frac{3}{5}[/tex]Where:
[tex]0You need to remember that, by definition:[tex]\theta=\cos ^{-1}(\frac{adjacent}{hypotenuse})[/tex]Therefore, applying this formula, you can find the measure of angles A and B:
[tex]A=\cos ^{-1}(\frac{5}{13})\approx67.38\text{\degree}[/tex][tex]B=\cos ^{-1}(\frac{3}{5})\approx53.13\text{\degree}[/tex](a) By definition:
[tex]\sin \mleft(A+B\mright)=sinAcosB+cosAsinB[/tex]Knowing that:
[tex]\sin \theta=\frac{opposite}{hypotenuse}[/tex]You can substitute the known values into the equation in order to find the opposite side for angle A:
[tex]\begin{gathered} \sin (67.38\text{\degree)}=\frac{opposite}{13} \\ \\ 13\cdot\sin (67.38\text{\degree)}=opposite \\ \\ opposite\approx12 \end{gathered}[/tex]Now you know that:
[tex]\sin A=\frac{12}{13}[/tex]Using the same reasoning for angle B, you get:
[tex]\begin{gathered} \sin (53.13\text{\degree)}=\frac{opposite}{5} \\ \\ 5\cdot\sin (53.13\text{\degree)}=opposite \\ \\ opposite\approx4 \end{gathered}[/tex]Now you know that:
[tex]\sin B=\frac{4}{5}[/tex]Substitute values into the Trigonometric Identity:
[tex]\begin{gathered} \sin (A+B)=sinAcosB+cosAsinB \\ \\ \sin (A+B)=(\frac{12}{13})(\frac{3}{5})+(\frac{5}{13})(\frac{4}{5}) \end{gathered}[/tex]Simplifying, you get:
[tex]\begin{gathered} \sin (A+B)=\frac{36}{65}+\frac{20}{65} \\ \\ \sin (A+B)=\frac{36+20}{65} \end{gathered}[/tex][tex]\sin (A+B)=\frac{56}{65}[/tex](b) By definition:
[tex]\sin \mleft(A-B\mright)=sinAcosB-cosAsinB[/tex]Knowing all the values, you get:
[tex]\begin{gathered} \sin (A-B)=(\frac{12}{13})(\frac{3}{5})-(\frac{5}{13})(\frac{4}{5}) \\ \\ \sin (A-B)=\frac{36-20}{65} \\ \\ \sin (A-B)=\frac{16}{65} \end{gathered}[/tex](c) By definition:
[tex]\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\cdot\tan B}[/tex]By definition:
[tex]\tan \theta=\frac{opposite}{adjacent}[/tex]Therefore, in this case:
- For angle A:
[tex]\tan A=\frac{12}{5}[/tex]- And for angle B:
[tex]\tan B=\frac{4}{3}[/tex]Therefore, you can substitute values into the formula and simplify:
[tex]\tan (A+B)=\frac{\frac{12}{5}+\frac{4}{3}}{1-(\frac{12}{5}\cdot\frac{4}{3})}[/tex][tex]\tan (A+B)=\frac{\frac{56}{15}}{1-\frac{48}{15}}[/tex][tex]\tan (A+B)=\frac{\frac{56}{15}}{-\frac{11}{5}}[/tex][tex]\tan (A+B)=-\frac{56}{33}[/tex](d) By definition:
[tex]\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\cdot\tan B}[/tex]Knowing all the values, you can substitute and simplify:
[tex]\tan (A-B)=\frac{\frac{12}{5}-\frac{4}{3}}{1+(\frac{12}{5}\cdot\frac{4}{3})}[/tex][tex]\tan (A-B)=\frac{\frac{16}{15}}{\frac{21}{5}}[/tex][tex]\tan (A-B)=\frac{16}{63}[/tex](e) Knowing that:
[tex]\sin (A+B)=\frac{56}{65}[/tex][tex]\tan (A+B)=-\frac{56}{33}[/tex]Remember the Quadrants:
By definition, in Quadrant II the Sine is positive and the Tangent is negative.
Since in this case, you found that the Sine is positive and the Tangent negative, you can determine that this angle is in the Quadrant II:
[tex]A+B[/tex]What is the seventy-seven is forty-six more than r
Answer: 77 = 46 + r, r = 31
Step-by-step explanation:
We will write an equation to represent this situation. Then, we will solve for r by isolating the variable.
Seventy-seven is forty-six more than r.
77 is forty-six more than r.
77 = forty-six more than r.
77 = 46 more than r.
77 = 46 + r
77 = 46 + r
(77) - 46 = (46 + r) - 46
31 = r
r = 31
Sara spent 35 minutes on math homework and 20 minutes on reading homework. Mia spent a total of 40
minutes on reading and math homework. How much longer did Sara spend on her homework than Mia?
Sara spent 15 minutes longer than (the difference is 15 min) Mia in her homework.
According to the question,
We have the following information:
Sara spent 35 minutes on math homework and 20 minutes on reading homework. Mia spent a total of 40 minutes on reading and math homework.
So, it means that the total time spent by Sara in her homework is:
35+20 minutes
55 minutes
So, the differences between their time spent in her homework (will give us the more time taken by Sara) is:
Time spent by Sara in her homework-time spent by Mia in her homework
(55-40) minutes
15 minutes
Hence, Sara spent 15 more minutes than Mia.
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I need help question 10 b and c
Part b.
In this case, we have the following function:
[tex]y=5(2.4)^x[/tex]First, we need to solve for x. Then, by applying natural logarithm to both sides, we have
[tex]\log y=\log (5(2.4^x))[/tex]By the properties of the logarithm, it yields
[tex]\log y=\log 5+x\log 2.4[/tex]By moving log5 to the left hand side, we have
[tex]\begin{gathered} \log y-\log 5=x\log 2.4 \\ \text{which is equivalent to} \\ \log (\frac{y}{5})=x\log 2.4 \end{gathered}[/tex]By moving log2.4 to the left hand side, we obtain
[tex]\begin{gathered} \frac{\log\frac{y}{5}}{\log2.4}=x \\ or\text{ equivalently,} \\ x=\frac{\log\frac{y}{5}}{\log2.4} \end{gathered}[/tex]Therfore, the answer is
[tex]f^{-1}(y)=\frac{\log\frac{y}{5}}{\log2.4}[/tex]Part C.
In this case, the given function is
[tex]y=\log _{10}(\frac{x}{17})[/tex]and we need to solve x. Then, by raising both side to the power 10, we have
[tex]\begin{gathered} 10^y=10^{\log _{10}(\frac{x}{17})} \\ \text{which gives} \\ 10^y=\frac{x}{17} \end{gathered}[/tex]By moving 17 to the left hand side, we get
[tex]\begin{gathered} 17\times10^y=x \\ or\text{ equivalently,} \\ x=17\times10^y \end{gathered}[/tex]Therefore, the answer is
[tex]f^{-1}(y)=17\times10^y[/tex]36. The widths of platinum samples manufactured at a factory are normally distributed, with a mean of 1.3 cm and a standard deviation of 0.3 cm. Find the z-scores that correspond to each of the following widths. Round your answers to the nearest hundredth, if necessary.(a) 1.7 cmz = (b) 0.9 cmz =
Part (a)
Using the formula for the z-scores and the information given, we have:
[tex]\begin{gathered} \text{ z-score=}\frac{\text{ data value }-\text{ mean}}{\text{ standard deviation }} \\ \text{ z-score=}\frac{1.7\text{ cm }-\text{ 1.3 cm}}{0.3\text{ cm}} \\ \text{ z-score=}\frac{0.4\text{ cm}}{0.3\text{ cm}}\text{ (Subtracting)} \\ \text{ z-score=1.33 (Dividing)} \\ \text{The z-score for 1.7 cm is 1.33 rounding to the nearest hundredth.} \end{gathered}[/tex]Part (b)
Using the formula for the z-scores and the information given, we have:
[tex]\begin{gathered} \text{ z-score=}\frac{\text{ data value }-\text{ mean}}{\text{ standard deviation }} \\ \text{z-score=}\frac{\text{ 0.9 cm }-1.3\text{ cm}}{\text{ 0.3 }}\text{ (Replacing the values)} \\ \text{z-score=}\frac{\text{ }-0.4}{\text{ 0.3 }}\text{ (Subtracting)} \\ \text{ z-score= }-1.33 \\ \text{The z-score for 0.9 cm is -1.33 rounding to the nearest hundredth.} \end{gathered}[/tex]I need these answers quickly. If I don't get them by midnight ill cry.
*Statistical question: Is the proportion of inner-city families living on a subsistence income: 20%? Two hundred families were randomly selected for the survey
and 38 were found to have income at the subsistence level. Use the formal critical value method at 5% level of significance.
List the assumptions pertaining to this procedure.
Since the critical value of the test is greater than the absolute value of the test statistic, there is not enough evidence to conclude that the proportion is different of 20%.
Hypothesis tested and critical valueAt the null hypothesis, it is tested if the proportion is of 20%, that is:
[tex]H_0: p = 0.2[/tex]
At the alternative hypothesis, it is tested if the proportion is different of 20%, hence:
[tex]H_1: p \neq 0.2[/tex]
We have a two-tailed test, as we are testing if the mean is different of a value, with a significance level of 0.05, hence the critical value is of:
|z| = 1.96.
Test statisticThe test statistic is given by the rule presented as follows:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
[tex]\overline{p}[/tex] is the sample proportion.p is the proportion tested at the null hypothesis.n is the sample size.In the context of this problem, the parameters are given as follows:
[tex]p = 0.2, n = 200, \overline{p} = \frac{38}{200} = 0.19[/tex]
Hence the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.19 - 0.2}{\sqrt{\frac{0.2(0.8)}{200}}}[/tex]
z = -0.35.
|z| < 1.96, hence there is not enough evidence to conclude that the proportion is different of 20%.
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write 2500g in appropriate prefix pls.
Answer: 2.5kg
Step-by-step explanation:
I am assuming you mean to simplify it. So 2.5kg
1g=1000kg
2500/1000=2.5
**Determine the x-value at which the-following function touches but does not cross the x-axis:3x^3- 182 + 27x
Okay, here we have this:
We need to identify the x-value at which the-following function touches but does not cross the x-axis in the following function: 3x^3- 18^2 + 27x. So, considering that if is a zero with even multiplicity, the graph touches the x-axis and bounces off of the axis. And if it is a zero with odd multiplicity, the graph crosses the x-axis at a zero.
According with this let's
what is 0.024 ÷ 0.231
Answer:
0.10389610389
Step-by-step explanation:
Hi!
I plugged it into a calculator:
0.024 ÷ 0.231 = 0.10389610389
Have a great day! :)
Use the change of base formula and a calculator to evaluate the logarithm
The change of base formula states that:
[tex]\log _bx=\frac{\ln x}{\ln b}[/tex]this means that we can caculate any logarithm using the natural logarithm if we make the quotient of the natural logarithm of the original value and the natural logarithm of the original base.
In this case we have:
[tex]\begin{gathered} x=14 \\ b=\sqrt[]{3} \end{gathered}[/tex]Then, using the change of base formula, we have:
[tex]\log _{\sqrt[]{3}}14=\frac{\ln 14}{\ln \sqrt[]{3}}[/tex]Once we have the expression we just evaluate the expression on the right to get the appoximation we need:
[tex]\log _{\sqrt[]{3}}14=\frac{\ln14}{\ln\sqrt[]{3}}\approx4.804[/tex]Use the information given to find the equation of the line using the point-slope formula (y-y_1=m(x-x_1)). Then convert your answer to slope-intercept form (y=mx+b).(0,3) with a slope of 4The point slope form is (y-Answer)=Answer(x-Answer)Converting it to slope intercept form gives us y=Answerx+Answer
we have
m=4
point (0,3)
y-y1=m(x-x1)
substitute given values
y-3=4(x-0) ----> equation in point slope formConvert to slope-intercept form
y=mx+b
y-3=4x
adds 3 both sides
y=4x+3 ----> equation in slope-intercept formFind 2 given that =−4/5 and < < 3/2
Find 2 given that =
−4/5 and < < 3/2
we know that
sin(2x) = 2 sin(x) cos(x)
so
step 1
Find the value of cos(x)
Remember that
[tex]\sin ^2(x)+\cos ^2(x)=1^{}[/tex]we have
sin(x)=-4/5
The angle x lies on III quadrant
that means
cos(x) is negative
substitute the value of sin(x)
[tex]\begin{gathered} (-\frac{4}{5})^2+\cos ^2(x)=1^{} \\ \\ \frac{16}{25}+\cos ^2(x)=1^{} \\ \\ \cos ^2(x)=1-\frac{16}{25} \\ \cos ^2(x)=\frac{9}{25} \\ \cos (x)=-\frac{3}{5} \end{gathered}[/tex]step 2
Find the value of sin(2x)
sin(2x) = 2 sin(x) cos(x)
we have
sin(x)=-4/5
cos(x)=-3/5
substitute
sin(2x)=2(-4/5)(-3/5)
sin(2x)=24/25(2i) - (11+2i) complex numbers
Write the equation of the circle centered at (−4,−2) that passes through (−15,19)
In this problem, we are going to find the formula for a circle from the center and a point on the circle. Let's begin by reviewing the standard form of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]The values of h and k give us the center of the circle, (h,k). The value r is the radius. We can begin by substituting the values of h and k into our formula.
Since the center is at (-4, -2), we have:
[tex]\begin{gathered} (x-(-4))^2+(y-(-2))^2=r^2 \\ (x+4)^2+(y+2)^2=r^2 \end{gathered}[/tex]Next, we can use the center and the given point on the circle to find the radius.
Recall that the radius is the distance from the center of a circle to a point on that circle. So, we can use the distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Let
[tex](x_1,y_1)=(-4,-2)[/tex]and let
[tex](x_2,y_2)=(-15,19)[/tex]Now we can substitute those values into the distance formula and simplify.
[tex]\begin{gathered} r=\sqrt{(-15-(-4))^2+(19-(-2))^2} \\ r=\sqrt{(-11)^2+(21)^2} \\ r=\sqrt{562} \end{gathered}[/tex]Adding that to our equation, we have:
[tex]\begin{gathered} (x+4)^2+(y+2)^2=(\sqrt{562})^2 \\ (x+4)^2+(y+2)^2=562 \end{gathered}[/tex]You want to build a sandbox that can hold50,445 cubic inches of sand. If the sandbox is to be59 in. long and57 in. wide, how tall will it need to be?
Volume of sandbox (to be built) = 50,445 cubic inches
A sandbox is the shape of a cuboid and is calculated by the formula
[tex]\text{volume = length }\cdot\text{ wi}\differentialD tth\text{ }\cdot\text{ height }\Rightarrow\text{ v = l }\cdot\text{ w }\cdot\text{ h}[/tex]Volume = Length * Width * Height
Volume = 50,445 cubic inches, Length = 59 in. Width = 57 in, Height = ?
50,445 = 59 * 57 * h
Make h the subject of the formula, we have:
h = 50445 / (59 * 57) = 15 in
help meeeeeeeeee pleaseee !!!!!
The value of the composite function is: (f o g)(2) = 33.
How to Find the Value of a Composite Function?To evaluate a composite function, take the following steps:
Step 1: Find the value of the inner function by substituting the value of x into the equation of the functionStep 2: Use the value of the output of the inner function as the input for the outer function and simplify to get the value of the composite function.Given the following:
f(x) = x² - 3x + 5
g(x) = -2x
Therefore,
(f o g)(2) = f(g(2))
Find the value of the inner function g(2):
g(2) = -2(2)
g(2) = -4
Find f(g(2)) by substituting x = -4 into the function f(x) = x² - 3x + 5:
(f o g)(2) = f(g(2)) = (-4)² - 3(-4) + 5
= 16 + 12 + 5
(f o g)(2) = 33
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Find the product. Write your answer in scientific notation. (6.5 X 10^8) X (1.4 x 10^-5) =
Evaluate the product of the expression.
[tex]\begin{gathered} (6.5\times10^8)\cdot(1.4\times10^{-5})=6.5\cdot1.4\times10^{8-5} \\ =9.1\times10^3 \end{gathered}[/tex]So answer is 9.1X10^3.
What is the first operation that should be performed to calculate (3 + 2) × 6÷5 - 4?
A) addition
B) division
C) subtraction
D) multiplication
Answer: A) addition
Step-by-step explanation:
because of BODMAS, you need to do the bracket first
Enter a rule for each function f and g, and then compare their domains, ranges, slopes, and y-intercepts.The function f(x) has a slope of -2 and has a y-intercept of 3. The graph shows the function g(x).
The rule of the function f(x) is : -2x + 3
To find the rule of the function g(x) let's calculate the slope of the line
[tex]m=\frac{y2-y1}{x2-x1}=\frac{-11-5}{4-0}=\frac{-16}{4}=-4[/tex]The slope of the line is -4 and the intercept is 5 ( from the graph).
The rule of the function g(x) is : -4x + 5
The domains of f(x) and g(x) is All real numbers, because there is not any number of x which doesn't have a corresponding y-coordinate.
The ranges of f(x) and g(x) is All real numbers, because there is not any number of y which doesn't have a corresponding x-coordinate.
The slope of f(x) is greater than g(x) (-2 is greater than -4)
The y-intercept of f(x) is less than the y-intercept of g(x).(3 is less than 5)
Find the volume of this triangular prism.Be sure to include the correct unit in your answer.8 cm7 cm→5 cm
The formula to find the volume of a triangular prism is the following:
[tex]V=\frac{1}{2}h\cdot b\cdot w[/tex]where:
h - height
b - base length
w - width
for this problem:
h = 8 cm
b = 5 cm
w = 7 cm
then
[tex]V=\frac{1}{2}8\cdot5\cdot7[/tex]solving this, we obtain that the volume of the triangular prism is 140 cm^3 or cubic centimeters
Draw the following vectors using the scale 1 cm = 50 km/h. Plant the tail at the origin. A. 200 km/h on a bearing of 020° B. 75 km/h S 10° W C. 350 km/h NE
Solution
a)
200 km/h on a bearing of 020°
Scale 1 cm = 50 km/h.
[tex]Length\text{ = }\frac{200}{50}\text{ = 4cm}[/tex]b)
B. 75 km/h S 10° W
[tex]Lenght\text{ = }\frac{75}{50}\text{ = 1.5cm}[/tex]C. 350 km/h NE
[tex]Length\text{ = }\frac{350}{50}\text{ = 7cm}[/tex]Four research teamed each used a different method to collect data on how fast a new strain of maize sprouts. Assume that they all agree on the sample size and the sample mean ( in hours). Use the (confidence level; confidence interval) pairs below to select the team that has the smallest sample standard deviation
We need to identify the team that has the smallest sample standard deviation.
In order to do so, we need to find the stand deviation of each experiment based on the confidence level and confidence interval of each of them.
A. A confidence level of 99.7% corresponds to a confidence interval of 3 standard deviations above and 3 standard deviations below the mean.
Thus, for the confidence interval 42 to 48, the mean is 45. And the standard deviation is given by:
[tex]\begin{gathered} 3\sigma=48-45=3 \\ \\ \sigma=\frac{3}{3} \\ \\ \sigma=1 \end{gathered}[/tex]B. A confidence level of 95% corresponds to a confident interval of 2 standard deviations above and 2 standard deviations below the mean.
Thus, for the confidence interval 43 to 47, the mean is 45. And the standard deviation is given by:
[tex]\begin{gathered} 2\sigma=47-45=2 \\ \\ \sigma=\frac{2}{2} \\ \\ \sigma=1 \end{gathered}[/tex]C. A confidence level of 68% corresponds to a confident interval of 1 standard deviation above and 1 standard deviation below the mean.
Thus, for the confidence interval 44 to 46, the mean is 45. And the standard deviation is given by:
[tex]\begin{gathered} \sigma=46-45 \\ \\ \sigma=1 \end{gathered}[/tex]D. Again, we have a confidence level of 95%, which corresponds to 2 standard deviations.
Thus, for the confidence interval 44 to 46, the mean is 45. And the standard deviation is given by:
[tex]\begin{gathered} 2\sigma=46-45=1 \\ \\ \sigma=\frac{1}{2} \\ \\ \sigma=0.5 \end{gathered}[/tex]Therefore, the team that has the smallest sample standard deviation is:
Answer
I don't understand please explain in simple words the transformation that is happeningwhat is the function notation
We have the next functions
[tex]f(x)=5^x^{}[/tex][tex]g(x)=2(5)^x+1[/tex]Function notation
[tex]g(x)=2(f(x))+1[/tex]Describe the transformation in words
we have 2 transformations, the 2 that multiplies the function f(x) means that we will have an expansion in the y axis by 2, the one means that we will have a shift up by one unit
Need help figuring out if the following is Real or Complex Question number 10
Explanation:
We have the expression:
[tex]i^3[/tex]where i represents the complex number i defined as follows:
[tex]i=\sqrt{-1}[/tex]To find if i^3 is real or complex, we represent it as follows:
[tex]i^3=i^2\times i[/tex]And we find the value of i^2 using the definition of i:
[tex]i^2=(\sqrt{-1})^2[/tex]Since the square root and the power of 2 cancel each other
[tex]\imaginaryI^2=-1[/tex]And therefore, using this value for i^2, we can now write i^3 as follows:
[tex]\begin{gathered} \imaginaryI^3=\imaginaryI^2\times\imaginaryI \\ \downarrow \\ \imaginaryI^3=(-1)\times\imaginaryI \end{gathered}[/tex]This simplifies to -i
[tex]\imaginaryI^3=-\imaginaryI^[/tex]Because -i is still a complex number, that means that i^3 is a complex number.
Answer: Complex
An observer in a lighthouse 350 ft above sea level observes two ships directly offshore. The angles of depression to the shops are 4 degree and 6.5 degree. How far apart are the ships?
The two ships are 1933.32 ft apart
Explanation:Given:
The height of the lighthouse = 350 ft
The angles of depression to the ships are 4 degree and 6.5 degree
To find:
the distance between the two ships
To determine the distance, we will use an illustration of the situation
First we will find the value of y as we need to know this value to get x
To get y, we will apply tan ratio (TOA)
[tex]\begin{gathered} tan\text{ 6.5\degree = }\frac{opposite}{adjacent} \\ opp\text{ = 350 ft} \\ adj\text{ = y} \\ tan\text{ 6.5\degree = }\frac{350}{y} \\ y(tan\text{ 6.5\degree\rparen= 350} \\ y\text{ = }\frac{350}{tan\text{ 6.5}} \\ y\text{ = 3071.9106 ft} \end{gathered}[/tex]Next is to find x using tan ratio (TOA):
[tex]\begin{gathered} angle\text{ = 4\degree} \\ tan\text{ 4\degree= }\frac{opposite}{adjacent} \\ \\ opposite\text{ = 350 ft} \\ adjacent\text{ = y + x} \\ tan\text{ 4\degree= }\frac{350}{y\text{ + x}} \end{gathered}[/tex][tex]\begin{gathered} tan\text{ 4 = }\frac{350}{3071.9106+x} \\ \frac{350}{tan\text{ 4}}\text{ = 3071.9106 + x} \\ 5005.2332\text{ = 3071.9106 + x} \\ x\text{ = 1933.3226} \\ \\ The\text{ ships are 1933.32 ft apart \lparen nearest hundredth\rparen} \end{gathered}[/tex]I need help finding 5 points. the vertex, 2 to the left of the vertex, and 2 points to the right of the vertex.
Let's convert the given equation first into a vertex form.
[tex]y=a(x-h)^2+k[/tex]where (h, k) is the vertex.
The vertex form of the equatio that we have is:
[tex]y=-2(x-0)^2+0[/tex]Hence, the vertex of the equation is at the origin (0, 0).
Since "a" is negative, our parabola is opening downward.
Let's identify two points to the left of the vertex. Let's say at x = -1. Replace "x" with -1 in the equation.
[tex]\begin{gathered} y=-2(-1)^2 \\ y=-2(1) \\ y=-2 \end{gathered}[/tex]Hence, we have a point to the left of the parabola at (-1, -2).
Let's say x = -2. Replace "x" with -2 in the equation.
[tex]\begin{gathered} y=-2(-2)^2 \\ y=-2(4) \\ y=-8 \end{gathered}[/tex]Hence, we also have another point to the left of the parabola at (-2, -8).
If our x is to the right of the vertex, say, x = 1. Replace "x" with 1 in the equation.
[tex]\begin{gathered} y=-2(1)^2 \\ y=-2(1) \\ y=-2 \end{gathered}[/tex]We have a point to the right of the parabola at (1, -2).
If x = 2, let's replace "x" with 2 in the equation.
[tex]\begin{gathered} y=-2(2)^2 \\ y=-2(4) \\ y=-8 \end{gathered}[/tex]Hence, we also have another point to the right of the parabola at (2, -8).
The graph of this equation is: