Answers
The derivative is the instantaneous rate of change of a function with respect to one of its variables. This is equivalent to finding the slope of the tangent line to the function at a point.
The function is given to be:
[tex]T(t)=Ate^{-kt}[/tex]where A and k are positive constants.
We can find the derivative of the function as follows:
[tex]T^{\prime}(t)=\frac{d}{dt}(Ate^{-kt})[/tex]Step 1: Pull out the constant factor
[tex]T^{\prime}(t)=A\cdot\frac{d}{dt}(te^{-kt})[/tex]Step 2: Apply the product rule
[tex]\frac{d(uv)}{dx}=u \frac{dv}{dx}+v \frac{du}{dx}[/tex]Let
[tex]\begin{gathered} u=t \\ v=e^{-kt} \\ \therefore \\ \frac{du}{dt}=1 \\ \frac{dv}{dt}=-ke^{-kt} \end{gathered}[/tex]Therefore, we have:
[tex]T^{\prime}(t)=A(t\cdot(-ke^{-kt})+e^{-kt}\cdot1)[/tex]Step 3: Simplify
[tex]T^{\prime}(t)=A(-kte^{-kt}+e^{-kt})[/tex]QUESTION A
At t = 0, the instantaneous rate of change is calculated to be:
[tex]\begin{gathered} t=0 \\ \therefore \\ T^{\prime}(0)=A(-k(0)e^{-k(0)}+e^{-k(0)}) \\ T^{\prime}(0)=A(0+e^0) \\ Recall \\ e^0=1 \\ \therefore \\ T^{\prime}(0)=A \end{gathered}[/tex]The rate of change is:
[tex]rate\text{ }of\text{ }change=A[/tex]QUESTION B
At t = 30, the instantaneous rate of change is calculated to be:
[tex]\begin{gathered} t=30 \\ \therefore \\ T(30)=A(-k(30)e^{-k(30)}+e^{-k(30)}) \\ T(30)=A(-30ke^{-30k}+e^{-30k}) \\ Collecting\text{ }common\text{ }factors \\ T(30)=Ae^{-30k}(-30k+1) \end{gathered}[/tex]The rate of change is:
[tex]rate\text{ }of\text{ }change=Ae^{-30k}(-30k+1)[/tex]QUESTION C
When the rate of change is equal to 0, we have:
[tex]0=A(-kte^{-kt}+e^{-kt})[/tex]We can make t the subject of the formula using the following steps:
Step 1: Apply the Zero Factor principle
[tex]\begin{gathered} If \\ ab=0 \\ a=0,b=0 \\ \therefore \\ -kte^{-kt}+e^{-kt}=0 \end{gathered}[/tex]Step 2: Collect common terms
[tex]e^{-kt}(-kt+1)=0[/tex]Step 3: Apply the Zero Factor Principle:
[tex]\begin{gathered} e^{-kt}=0 \\ \ln e^{-kt}=\ln0 \\ -kt=\infty \\ t=\infty \end{gathered}[/tex]or
[tex]\begin{gathered} -kt+1=0 \\ -kt=-1 \\ t=\frac{-1}{-k} \\ t=\frac{1}{k} \end{gathered}[/tex]The time will be:
[tex]t=\frac{1}{k}[/tex]Related Questions
Can you please solve the last question… number 3! Thanks!
Answers
Let us break the shape into two triangles and solve for the unknowns.
The first triangle is shown below:
We will use the Pythagorean Theorem defined to be:
[tex]\begin{gathered} c^2=a^2+b^2 \\ where\text{ c is the hypotenuse and a and b are the other two sides} \end{gathered}[/tex]Therefore, we can relate the sides of the triangles as shown below:
[tex]25^2=y^2+16^2[/tex]Solving, we have:
[tex]\begin{gathered} y^2=25^2-16^2 \\ y^2=625-256 \\ y^2=369 \\ y=\sqrt{369} \\ y=19.2 \end{gathered}[/tex]Hence, we can have the second triangle to be:
Applying the Pythagorean Theorem, we have:
[tex]22^2=x^2+19.2^2[/tex]Solving, we have:
[tex]\begin{gathered} 484=x^2+369 \\ x^2=484-369 \\ x^2=115 \\ x=\sqrt{115} \\ x=10.7 \end{gathered}[/tex]The values of the unknowns are:
[tex]\begin{gathered} x=10.7 \\ y=19.2 \end{gathered}[/tex]I need help with this question... the correct answer choice
Answers
Reflection over the x-axis:
(x,y)--->(x, -y)
and the question is what is not a reflection across the x-axis.
so,
the correct option is D which is:
R'(-9, 4) ----> R'(9, -4)
Because it is a reflection over the y-axis.
-27\sqrt(3)+3\sqrt(27), reduce the expression
Answers
Explanation
[tex]-27\sqrt[]{3}+3\sqrt[]{27}[/tex]Step 1
Let's remember one propertie of the roots
[tex]\sqrt[]{a\cdot b}=\sqrt[]{a}\cdot\sqrt[]{b}[/tex]hence
[tex]\sqrt[]{27}=\sqrt[]{9\cdot3}=\sqrt[]{9}\cdot\sqrt[]{3}=3\sqrt[]{3}[/tex]replacing in the expression
[tex]\begin{gathered} -27\sqrt[]{3}+3\sqrt[]{27} \\ -27\sqrt[]{3}+3(3\sqrt[]{3}) \\ -27\sqrt[]{3}+9\sqrt[]{3} \\ (-27+9)\sqrt[]{3} \\ -18\sqrt[]{3} \end{gathered}[/tex]therefore, the answer is
[tex]-18\sqrt[]{3}[/tex]I hope this helps you
What is 3 +4.3+45?A4늘OB.B. 7O. 8○ D. 12
Answers
solution
[tex]3+4\frac{1}{3}=7\frac{1}{3}[/tex]answer: B
5-3/2x>1/3what is x?
Answers
Coco, this is the solution to the inequality:
5 - 3x/2 ≥ 1/3
Subtracting 5 at both sides:
5 - 3x/2 - 5 ≥ 1/3 - 5
-3x/2 ≥ 1/3 - 15/3
-3x/2 ≥ -14/3
LCD (Least Common Denominator) between 2 and 3 : 6
-9x/6 ≥ -28/6
Dividing by -9/6 at both sides:
-9x/6 / -9/6 ≥ -28/6 / -9/6
x ≥ 28/9
In consequence, the correct answer is C. x ≥ 28/9
Surface are of the wood cube precision =0.00The weight of the woo cube precision =0.00 The volume was 42.87 in3
Answers
Given:
The volume of the cube is 42.87 cubic inches.
The volume of a cube is given as,
[tex]\begin{gathered} V=s^3 \\ 42.87=s^3 \\ \Rightarrow s=3.5 \end{gathered}[/tex]The surface area of a cube is,
[tex]\begin{gathered} SA=6s^2 \\ SA=6\cdot(3.5)^2 \\ SA=73.5 \end{gathered}[/tex]Answer: the surface area is 73.5 square inches ( approximately)
Which of the following shows the expansion of sum from n equals 0 to 4 of 2 minus 5 times n ?
(−18) + (−13) + (−8) + (−3) + 0
(−3) + (−8) + (−13) + (−18) + (−23)
2 + (−3) + (−8) + (−13) + (−18)
2 + 7 + 12 + 17 + 22
Answers
The option that indicates the required sum when n equals 0 to 4 of 2 minus 5 times n, is 2 + (−3) + (−8) + (−13) + (−18) (Option C)
What is the Sum of sequences?The sum of the terms of a sequence is called a series.
From the given sum of a sequence, we are to find the sum of the given sequence from n = 0 to n = 4
When n = 0
a(0) = 2 - 5(0)
a(0) = 2 - 0
a(0) = 2
When n = 1
a(1) = 2 - 5(1)
a(1) = 2 -5
a(1) = -3
When n = 2
a(2) = 2 - 5(2)
a(2) = 2 - 10
a(2) = -8
When n = 3
a(3) = 2 - 5(3)
a(3) = 2 - 15
a(3) = -13
When n = 4
a(4) = 2 - 5(4)
a(4) = 2 - 20
a(4) = -18
Hence the required sum is 2 + (−3) + (−8) + (−13) + (−18)
Learn more about sum of sequences:
https://brainly.com/question/24160061
#SPJ1
The required sum is 2 + (−3) + (−8) + (−13) + (−18) when n equals 0 to 4 of 2 minus 5 times n, which is the correct answer that would be an option (C).
The given expression is (2 - 5n)
We to determine the sum of the given sequence from n = 0 to n = 4
Let the required sum is T₀ + T₁ + T₂ + T₃ + T₄
Substitute the value of n = 0 in the expression (2 - 5n) to get T₀
⇒ T₀ = 2 - 5(0) = 2 - 0 = 2
Substitute the value of n = 1 in the expression (2 - 5n) to get T₁
⇒ T₁ = 2 - 5(1) = 2 -5 = -3
Substitute the value of n = 2 in the expression (2 - 5n) to get T₂
⇒ T₂ = 2 - 5(2) = 2 - 10 = -8
Substitute the value of n = 3 in the expression (2 - 5n) to get T₃
⇒ T₃ = 2 - 5(3) = 2 - 15 = -13
Substitute the value of n = 4 in the expression (2 - 5n) to get T₄
⇒ T₄ = 2 - 5(4) = 2 - 20 = -18
Therefore, the required sum is 2 + (−3) + (−8) + (−13) + (−18)
Learn more about the sequences here:
brainly.com/question/21961097
#SPJ1
9. SAILING The sail on Milton's schooner is the shape of a 30°-60°-90°triangle. The length of the hypotenuse is 45 feet. Find the lengths of thelegs. Round to the nearest tenth.
Answers
The triangle is shown below:
Notice how this is an isosceles triangle.
We can find the lengths of the hypotenuse by using the trigonometric functions:
[tex]\sin \theta=\frac{\text{opp}}{\text{hyp}}[/tex]Then we have:
[tex]\begin{gathered} \sin 45=\frac{21}{hyp} \\ \text{hyp}=\frac{21}{\sin 45} \\ \text{hyp}=29.7 \end{gathered}[/tex]Therefore the hypotenuse is 29.7 ft.
Lne segment AC and BD are parallel, what are the new endpoints of the line segments AC and BD if the parallel lines are reflected across the y-axis?
Answers
Given a point P = (x, y) a reflection P' alongside the y axis of that point follows the rule:
[tex]P=(x,y)\Rightarrow P^{\prime}=(-x,y)[/tex]We need to multiply the x coordinates of the points by (-1)
The cordinates of the points in the problem are:
A = (2, 5)
B = (2, 4)
C = (-5, 1)
D = (-5, 0)
Then the endpoints of the reflection over the y axis are:
A' = (-2, 5)
B' = (-2, 4)
C' = (5, 1)
D' = (5, 0)
Which is the second option.
Suppose the coordinate of p=2 and PQ=8. Whare are the possible midpoints for PQ?
Answers
The midpoint for segment PQ can be calculated as:
[tex]\frac{P+Q}{2}[/tex]Then, the midpoint of PQ is:
[tex]\frac{2\text{ + Q}}{2}=1+0.5Q[/tex]Additionally, PQ can be calculated as:
[tex]PQ=\left|Q-P\right|[/tex]So:
[tex]\begin{gathered} \left|Q-P\right|=8 \\ \left|Q-2\right|=8 \end{gathered}[/tex]It means that:
[tex]\begin{gathered} Q-2=8\text{ or } \\ 2\text{ - Q = 8} \end{gathered}[/tex]Solving for Q, we get:
Q = 8 + 2 = 10 or Q = 2 - 8 = -6
Finally, replacing these values on the initial equation for the midpoint, we get:
If Q = 10, then:
midpoint = 1 + 0.5(10) = 1 + 5 = 6
If Q = -6, then:
midpoint = 1 + 0.5(-6) = 1 - 3 = -2
The possible midpoints for PQ are 6 and -2
Find the sum of the arithmetic series -1+ 2+5+8+... where n=7.A. 56B. 184C. 92D. 380Reset Selection
Answers
The arithmetic series is:
-1 + 2 + 5 + 8 + .....
The first term, a = -1
The common difference, d = 2 - (-1)
d = 3
The number of terms, n = 7
Find the sum of the arithmetic series below
[tex]\begin{gathered} S_n=\frac{n}{2}[2a+(n-1)d] \\ \\ S_7=\frac{7}{2}[2(-1)+(7-1)(3)] \\ \\ S_7=\frac{7}{2}(-2+18) \\ \\ S_7=\frac{7}{2}(16) \\ \\ S_7=56 \end{gathered}[/tex]Therefore, the sum of the arithmetic series = 56
Use an inequality to represent the corresponding Celsius temperature that is at or below 32° F.
Answers
C ≤ 0
Explanations:The given equation is:
[tex]F\text{ = }\frac{9}{5}C\text{ + 32}[/tex]Make C the subject of the equation
[tex]\begin{gathered} F\text{ - 32 = }\frac{9}{5}C \\ 9C\text{ = 5(F - 32)} \\ C\text{ = }\frac{5}{9}(F-32) \end{gathered}[/tex]At 32°F, substitute F = 32 into the equation above to get the corresponding temperature in °C
[tex]\begin{gathered} C\text{ = }\frac{5}{9}(32-32) \\ C\text{ = }\frac{5}{9}(0) \\ C\text{ = 0} \end{gathered}[/tex]The inequality representing the corresponding temperature that is at or below 32°F is C ≤ 0
Find the exact solution to the exponential equation. (No decimal approximation)
Answers
Let's solve the equation:
[tex]\begin{gathered} 54e^{3x+3}=16 \\ e^{3x+3}=\frac{16}{54} \\ e^{3x+3}=\frac{8}{27} \\ \ln e^{3x+3}=\ln (\frac{8}{27}) \\ 3x+3=\ln (\frac{2^3}{3^3}) \\ 3x+3=\ln (\frac{2}{3})^3 \\ 3x+3=3\ln (\frac{2}{3}) \\ 3x=-3+3\ln (\frac{2}{3}) \\ x=-1+\ln (\frac{2}{3}) \\ x=-1+\ln 2-\ln 3 \end{gathered}[/tex]Therefore the solution of the equation is:
[tex]x=-1+\ln 2-\ln 3[/tex]I NEED HELP
5C/2 = 20
Answers
you would have to do this backwards
20 times 2 would remove the /2
5c=40
40 divided by 5
is
8
C=8
The slope and y-intercept of the relation represented by the equation 12x-9y+12=0 are:
Answers
12x - 9y +12 =0
To find the slope and y intercept, we want to put the equation in slope intercept form
y = mx+b where m is the slope and b is the y intercept
Solve the equation for y
Add 9y to each side
12x - 9y+9y +12 =0+9y
12x+12 = 9y
Divide each side by 9
12x/9 +12/9 = 9y/9
4/3 x + 4/3 = y
Rewriting
y = 4/3x + 4/3
The slope is 4/3 and the y intercept is 4/3
I inserted a picture of the question can you please hurry
Answers
Given:
[tex](-2,-5)\text{ and (}1,4)\text{ are given points.}[/tex][tex]\begin{gathered} \text{Slope}=\frac{y_2-y_1}{x_2-x_1} \\ \text{Slope}=\frac{4+5}{1+2} \\ \text{Slope}=\frac{9}{3} \\ \text{Slope}=3 \end{gathered}[/tex]The cost, c(x) in dollars per hour of running a trolley at an amusement park is modelled by the function [tex]c(x) = 2.1x {}^{2} - 12.7x + 167.4[/tex]Where x is the speed in kilometres per hour. At what approximate speed should the trolley travel to achieve minimum cost? A. About 2km/h B about 3km/h C about 4km/D about 5km/hr
Answers
The equation is modelled by the function,
c(x) = 2.1x^2 - 12.7x + 167.4
The general form of a quadratic equation is expressed as
ax^2 + bx + c
The given function is quadratic and the graph would be a parabola which opens upwards because the value of a is positive
Since x represents the speed, the speed at which the he
Please help me no other tutor could or understand it
Answers
We must find the equation that models the amount of medication in the bloodstream as a function of the days passed from the initial dose. The initial dose is a and we are going to use x for the number of days and M for the amount of mediaction in the bloodstream. We are going to model this using an exponential function which means that the variable x must be in the exponent of a power:
[tex]M(x)=a\cdot b^x[/tex]We are told that the half-life of the medication is 6 hours. This means that after 6 hours the amount of medication in the bloodstream is reduced to a half. If the initial dose was a then the amount after 6 hours has to be a/2. We are going to use this to find the parameter b but first we must convert 6 hours into days since our equation works with days.
Remember that a day is composed of 24 hours so 6 hours is equivalent to 6/24=1/4 day. This means that the amount of medication after 1/4 days is the half of the initial dose. In mathematical terms this means M(1/4)=M(0)/2:
[tex]\begin{gathered} \frac{M(0)}{2}=M(\frac{1}{4}) \\ \frac{a\cdot b^0}{2}=a\cdot b^{\frac{1}{4}} \\ \frac{a}{2}=a\cdot b^{\frac{1}{4}} \end{gathered}[/tex]We can divide both sides of this equation by a:
[tex]\begin{gathered} \frac{\frac{a}{2}}{a}=\frac{a\cdot b^{\frac{1}{4}}}{a} \\ \frac{1}{2}=b^{\frac{1}{4}} \end{gathered}[/tex]Now let's raised both sides of this equation to 4:
[tex]\begin{gathered} (\frac{1}{2})^4=(b^{\frac{1}{4}})^4 \\ \frac{1}{2^4}=b^{\frac{1}{4}\cdot4} \\ b=\frac{1}{16} \end{gathered}[/tex]Which can also be written as:
[tex]b=16^{-1}[/tex]Then the equation that models how much medication will be in the bloodstream after x days is:
[tex]M(x)=a\cdot16^{-x}[/tex]Using this we must find how much medication will be in the bloodstream after 4 days for an initial dose of 500mg. This basically means that a=500mg, x=4 and we have to find M(4):
[tex]M(4)=500mg\cdot16^{-4}=0.00763mg[/tex]So after 4 days there are 0.00763 mg of medication in the bloodstream.
Now we have to indicate how much more medication will be if the initial dose is 750mg instead of 500mg. So we take a=750mg and x=4:
[tex]M(4)=750mg\cdot16^{-4}=0.01144mg[/tex]If we substract the first value we found from this one we obtained the required difference:
[tex]0.01144mg-0.00763mg=0.00381mg[/tex]So the answer to the third question is 0.00381mg.
Need answer if you could show work would be nice
Answers
x + 3 = 0 when x = -3
x^2 - 6x + 13 = 0
x^2 - 6x + 9 + 4 = 0
(x - 3)^2 = -4
This shows x^2 - 6x + 13 = 0 has no real roots
The function only has one real root at x = -3
write the equation for a quadratic function in vertex form that opebs down shifts 8 units to the left and 4 units down .
Answers
STEP - BY STEP EXPLANATION
What to find?
Equation for a quadratic equation.
Given:
Shifts 8 unit to the left.
4 units down
Step 1
Note the following :
• The parent function of a quadratic equation in general form is given by;
[tex]y=x^2[/tex]• If f(x) shifts q-units left, the f(x) becomes, f(x+q)
,• If f(x) shift m-units down, then the new function is, f(x) -m
Step 2
Apply the rules to the parent function.
8 units to the left implies q=8
4 units down implies m= 4
[tex]y=(x+8)^2-4[/tex]ANSWER
y= (x+8)²- 4
Using the distributive property, show how to decompose 8 * 78
Answers
Given any three numbers a, b, and c.
By the distributive law, we must have:
a x (b + c) = (a x b) + (a x c)
Now to find 8 x78
8 x 78 = 8 x (70 + 8) = (8 x 70) + (8 x 8) = 560 + 64 = 624
(I don't know if there are tutors here right now at this time but it's worth a try.) Please help me I really really don't understand this, it's going to take me a while to understand this. X(
Answers
by the distributive law x(y+z)=zy+xz, we have
[tex]\begin{gathered} 3b+3(5)=4(2b)-4(5) \\ 3b+15=8b-20 \end{gathered}[/tex]Then we use the properties of inequalities, we can switch both sides, and if we add or multiply something on both sides the equality remains
[tex]\begin{gathered} 3b+15=8b-20 \\ \end{gathered}[/tex]we want the variables and the numbers without variables to be in different side, so, first we add 20 to both sides, note that the -20 will be cancelled
[tex]\begin{gathered} 3b+15+20\text{ = 8b-20+20} \\ 3b+15+20=8b \end{gathered}[/tex]we want to left all the numbers with variable on the right side so we substract 3b (add -3b) to both sides. Same as before, the 3b will be cancellated (we can change the order in the sum)
[tex]\begin{gathered} -3b+3b+15+20=-3b+8b \\ 15+20=8b-3b \end{gathered}[/tex]of course, you're welcome
I was asking if you have understood my explanation so far
tell me
it doesn't matter the order, in fact, when you get used to the method you can work with both at the same time
any other question?
yes, you could substrac 3b first
For example
[tex]\begin{gathered} 2+3x=6-x \\ 2+3x+x=6-x+x \\ 2+3x+x=6 \\ -2+2+3x+x=-2+6 \\ 3x+x=6-2 \\ 4x=4 \\ \end{gathered}[/tex]sadly I will need to leave since my shift is over, but if you ask another question one of my partners will help you
Have a nice evening!!!!
then we add like terms and switch both sides
[tex]5b=35[/tex]And then we multiply by 1/5 both sides
[tex]\begin{gathered} 5\frac{1}{5}b=\frac{35}{5} \\ b=\frac{35}{5} \\ b=7 \end{gathered}[/tex]Rewrite the fraction with a rational denominator:
[tex]\frac{1}{\sqrt{5} +\sqrt{3} -1}[/tex]
Give me a clear and concise explanation (Step by step)
I will report you if you don't explain
Answers
The expression with rational denominator is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
How to rewrite the fraction?From the question, the fraction is given as
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1}[/tex]
To rewrite the fraction with a rational denominator, we simply rationalize the fraction
When the fraction is rationalized, we have the following equation
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{1}{\sqrt 5 + \sqrt{3} - 1} \times \frac{\sqrt 5 - \sqrt{3} + 1}{\sqrt 5 - \sqrt{3} + 1}[/tex]
Evaluate the products in the above equation
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{(\sqrt 5)^2 - (\sqrt{3} + 1)^2}[/tex]
This gives
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{5 - 10 - 2\sqrt 3}[/tex]
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3}[/tex]
Rationalize again
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{\sqrt 5 - \sqrt{3} + 1}{- 5 - 2\sqrt 3} \times \frac{- 5+2\sqrt 3}{- 5 +2\sqrt 3}[/tex]
This gives
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{(-5)^2 - (2\sqrt 3)^2}[/tex]
So, we have
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{25 -12}[/tex]
Evaluate
[tex]\frac{1}{\sqrt 5 + \sqrt{3} - 1} = \frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
Hence, the expression is [tex]\frac{(\sqrt 5 - \sqrt{3} + 1)(- 5+2\sqrt 3)}{13}[/tex]
Read more about rational expressions at
https://brainly.com/question/20400557
#SPJ1
ExpenseYearly costor rateGasInsuranceWhat is the cost permile over the course ofa year for a $20,000 carthat depreciates 20%,with costs shown in thetable, and that hasbeen driven for 10,000miles?$425.00$400.00$110,00$100.0020%OilRegistrationDepreciationB. $4.10 per mileA. $1.10 per mileC. $0.25 per mileD. $0.50 per mile
Answers
Step 1:
Find the depreciation
[tex]\begin{gathered} \text{Depreciation = 20\% of \$20,000} \\ \text{Depreciation = }\frac{20}{100}\text{ }\times\text{ \$20000} \\ \text{Depreciation = \$4000} \end{gathered}[/tex]Step 2:
Total cost = $425 + $400 + $110 + $100 + $4000
Total cost = $5035
Final answer
[tex]\begin{gathered} \text{Cost per mile = }\frac{5035}{10000} \\ \text{Cost per mile = \$0.5035} \\ \text{Cost per mile = \$0.50} \end{gathered}[/tex]Option D $0.50 per mile
The answer is 57.3 provided by my teacher, I need help with the work
Answers
Apply the angles sum property in the triangle ABC,
[tex]62+90+\angle ACB=180\Rightarrow\angle ACB=180-152=28^{}[/tex]Similarly, apply the angles sum property in triangle BCD,
[tex]20+90+\angle BCD=180\Rightarrow\angle BCD=180-110=70[/tex]From triangle ABC,
[tex]BC=AC\sin 62=30\sin 62\approx26.5[/tex]From triangle BDC,
[tex]BD=BC\cos 20=26.5\cos 20\approx24.9[/tex]Now, consider that,
[tex]\angle BDE+\angle BDC=180\Rightarrow\angle BDE+90=180\Rightarrow\angle BDE=90[/tex]So the triangle BDE is also a right triangle, and the trigonometric ratios are applicable.
Solve for 'x' as,
[tex]x=\tan ^{-1}(\frac{BD}{DE})=\tan ^{-1}(\frac{24.9}{16})=57.2764\approx57.3[/tex]Thus, the value of the angle 'x' is 57.3 degrees approximately.ang
is 2÷2 4 or am I wrong
Answers
2/2 = 1
The answer would be 1
Gloria's teacher asks her to draw a triangle with a 90° angle and a 42° angle.How many unique triangles can Gloria draw that meet her teacher's requirements?AOne unique triangle can be drawn because the third angle must measure 48º.BNo unique triangle can be drawn because the teacher only gave the measures of two angles.СInfinitely many unique triangles can be drawn because the side lengths of the triangles can be different sizes.DThere is not enough information to determine how many unique triangles can be drawn.
Answers
SOLUTION
Sum of angles in a triangle must be equal to 180°
So, since one of the angle measures 90° and the other is 42°, then
90 + 42 + y = 180°, where y is the third angle
So, 132 + y = 180
y = 180 - 132 = 48°.
Therefore, one unique triangle can be drawn because the third angle must measure 48º.
Option A is the correct answer.
A square has a perimeterof 8,000 centimeters. Whatis the length of each side ofthe of the square inmeters?
Answers
Answer:
20 meters
Explanation:
The formula for calculating the perimeter of a square is expressed as
perimeter = 4s
where
s is the length of each side of the square
From the information given,
perimeter = 8,000 centimeters
Recall,
1 cm = 0.01 m
8000cm = 8000 x 0.01 = 80 m
Thus,
80 = 4s
s = 80/4
s = 20
The length of each side of the square is 20 meters
PLEASE HELP AS SOON AS POSSIBLE PLEASE!! ( one question, can whole numbers be classified as integers and rational numbers)
Answers
ANSWER
G. 10 and -2 only
EXPLANATION
-5/4 is a fraction that can't be simplified. Therefore it is not an integer.
1.25 has decimals, so it is not an integer either.
10 and -2 are are integers.
What would the answer be?
Nvm, I got it wrong
Answers
Applying the definition of similar triangles, the measure of ∠DEF = 85°.
What are Similar Triangles?If two triangles are similar, then their corresponding angles are all equal in measure to each other.
In the image given, since E and F are the midpoint of both sides of triangle BCD, then it follows that triangles BCD and EFD are similar triangles.
Therefore, ∠DBC ≅ ∠DEF
m∠DBC = m∠DEF
Substitute
4x + 53 = -6x + 133
4x + 6x = -53 + 133
10x = 80
10x/10 = 80/10 [division property of equality]
x = 8
Measure of ∠DEF = -6x + 133 = -6(8) + 133
Measure of ∠DEF = 85°
Learn more about similar triangles on:
https://brainly.com/question/14285697
#SPJ1