Answer:
[tex]f\lbrack h(7)\rbrack\text{ = 99}[/tex]Explanation:
Given the functions:
[tex]\begin{gathered} f(x)=2x-1 \\ h(x)=x^2+1 \end{gathered}[/tex]We want to find:
[tex]f\lbrack h(7)\rbrack[/tex]First of all, we need to find:
[tex]f\lbrack h(x)\rbrack[/tex]This is done by inserting the value of h(x) into f(x)
So, we have:
[tex]\begin{gathered} f\lbrack h(x)\rbrack=2(x^2+1)-1 \\ =2x^2+2-1 \\ =2x^2+1 \end{gathered}[/tex]Substituting 7 for x in f[h(x)], we have f[h(7)]
[tex]\begin{gathered} f\lbrack h(7)\rbrack=2(7^2)+1 \\ =2(49)+1 \\ =98+1 \\ =99 \end{gathered}[/tex]Which is what we are looking for.
What is the area of the composite figure? 9 in. 12 in. 24 in 20 in 12 in 15 in 30 in. O 1,182 square inches O 1,236 square inches O 978 square inches O 924 square inches
Given data:
The given figure is shown.
The area of the given figure is,
[tex]\begin{gathered} A=(24\text{ in)}(30\text{ in)+}\frac{1}{2}(24\text{ in)(9 in)+}\frac{1}{2}(15\text{ in)}(20\text{ in)} \\ =720\text{ sq-inches+108 sq-inches+150 sq-inches} \\ =978\text{ sq-inches} \end{gathered}[/tex]Thus, the area of the composite figure is 978 sq-inches.
(6 x 10^-2)(1.5 x 10^-3 + 2.5 x 10^-3)1.5 x 10^3
Given the expression:
[tex]\left(6*10^{-2}\right)\left(1.5*10^{-3}+2.5*10^{-3}\right)1.5*10^3[/tex]Let's simplify the expression.
To simplify the expression, we have:T
[tex]\begin{gathered} (6*10^{-2})(1.5*10^{-3}+2.5*10^{-3})1.5*10^3 \\ \\ =(6*10^{-2})(4.0*10^{-3})1.5*10^3 \\ \\ =(6*4.0*10^{-2-3})1.5*10^3 \\ \\ =(24.0*10^{-5})1.5*10^3 \end{gathered}[/tex]Solving further:
Apply the multiplication rule for exponents.
[tex]\begin{gathered} 24.0*1.5*10^{-5+3} \\ \\ =36*10^{-2} \\ \\ =0.36 \end{gathered}[/tex]ANSWER:
[tex]0.36[/tex]select the reason that best supports statement 6 in the given proof please help me image attached
Answer:
B. Distributive Property
Step-by-step explanation:
You want to know the reason in the proof that best supports the transition from 5. 99-3x = 12(x+2) to 6. 99-3x = 12x+24.
TransformationYou will notice that in the transition from
5. 99-3x = 12(x+2)
to
6. 99-3x = 12x+24
the expression 12(x+2) has been replaced by the expression 12x+24.
Distributive propertyThe property of addition and multiplication that makes it true that ...
12(x +2) = 12x +24
is the distributive property of multiplication over addition. That property tells you that parentheses can be eliminated by multiplying each of the terms inside by the factor outside.
Find the first three terms of this sequence Un=5n-2n3.
The first three terms of the sequence defined by the formula; Un=5n-2n³ as in the task content are; 3, -6 and -39 respectively.
What are the first three terms of the sequence given by the formula; Un=5n-2n³?It follows from the task content that the first three terms of the sequence defined by the formula be determined.
On this note, it follows that the first three terms are at; n = 1, n = 2 and n = 3 respectively.
Hence we have;
1st term; U(1) = 5(1) - 2(1)³ = 3.2nd term; U(2) = 5(2) - 2(2)³ = -6.3rd term; U(3) = 5(3) - 2(3)³ = -39.Hence, the first three terms are; 3, -6 and -39.
The first three terms of the sequence are as listed above.
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(1 point) For each trigonometric expression A,B,C,D, E, choose the expression from 1,2,3,4,5 that completes a fundamental identity. Enter the appropriate letter (A,B,C,D, or E) in each blank.
Answer:
Step-by-step explanation:
I would recommend looking up the magic trig hexagon, it has all of these identities and more within it.
1 - this corresponds with C as sin^2(x)+cos^2(x)=1
1-cos^2(x) - this corresponds with A, using the identity from number 1, we can rewrite it in the form sin^2(x)=1-cos^2(x)
cot(x) - for this it is important to know that cotangent is the inverse of tangent. Since tan(x)=sin(x)/cos(x), cot=cos(x)/sin(x) which is B.
sec^2(x) - much like the cos and sin pythagorean identity, sec and tan are related. sec^2(x)=tan^2(x)+1 which is answer choice E.
tan(x) - this is sin(x)/cos(x), choice D.
the radius of the circle is 5 inches. what is the area?give the exact answer in simplest form.
The area is 25π square inches
Explanation:Given a radius, r = 5 in.
The area of a circle is given by the formula:
[tex]A=\pi r^2[/tex]Substituting the value of r, we have:
[tex]A=\pi(5^2)=25\pi[/tex]The area is 25π square inches
Consider the following functions. Find the domain. Express your answer in interval notation.
Explanation:
[tex]\begin{gathered} f(x)\text{ = - }\sqrt[]{6-x} \\ g(x)\text{ = 4 - x} \\ (g\text{ - f)(x) = g(x) - f(x)} \end{gathered}[/tex][tex]\begin{gathered} (g\text{ -f)(x) = }4-\text{ x - (-}\sqrt[]{6\text{ - x}}) \\ (g\text{ -f)(x) = 4 - x + }\sqrt[]{6-x} \end{gathered}[/tex][tex]undefined[/tex]7. Reflect AABC over the y-axis, translate by (2, -1), and rotate the result 180° counterclockwise aboutthe origin. Plot AA'B'C' on the grid below. (1 point)tyTransformation rule:420А,PreimageABCImage A'B'CImage A"B"C"Image A'B'C'-22,-12,44, 2lifelongGeometry ACredit 2L4L - Geometry A (2020)Page 57
Reflection rule over y - axis is given as
(x , y) ------------ (-x, y)
This implies that the y - axis will remain the same and the x - axis will be negated
Pre image ABC at point (2, -1)
The reflection over y - axis will be
ABC A'B'C' A''B''C'' A'''B'''C'''
(2, -1) -----------------(-2, -1) ----------------------(2, -1) -------------------(-2, -1)
ABC A'B'C' A''B''C'' A'''B'''C'''
(2, -4) (-2, -4) (2, -4) (-2, -4)
A company borrows $13,000 at 5% for 90 days. Find (a) the amount of interest due and (b) the total amount that must be paid after 90 days. (a) The interest due is $ (Simplify your answer. Do not round until the final step. Then round to the nearest cent as needed.)
We have to use the simple interest formula
[tex]I=P\times r\times t[/tex]Where P = 13,000, r = 5% (0.05), t = 90 (0.25 years). Let's replace these values to find the interest
[tex]I=13,000\times0.05\times0.25=162.50[/tex](a) The amount of interest is $162.50.(b) The total amount that must be paid after 90 days is $13,162.50.Because we have to add the total interest with the amount borrowed.
A square room has a floor area of 49 square meters. The height of the room is 8 meters. What is the total area of all four walls?
The total area of all four walls is 224 square meters.
According to the question,
We have the following information:
A square room has a floor area of 49 square meters.
So, we have:
Area of square = 49 square meters
Side*side = 49
Side = [tex]\sqrt{49}[/tex] m
Side of the square = 7 m
Now, the side of the floor will be the width of the wall.
So, we have the width of the wall = 7 m.
The height of the room is 8 meters.
It means that the height of the wall is 8 m.
Area of 1 rectangular wall = length*width
Area of wall = 8*7
Area of 1 wall = 56 square meters
Now, the are of 4 walls will be (4*56) square meters or 224 square meters.
Hence, the total are of all four walls is 224 square meters.
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in triangle ABC, point E (5, 1.5) is the circumcenter, point He (4.3, 2.3) is the incente, and point I (3.6, 2.6) is the centroid.what is the approximate length of the radius that circumscribes triangle ABC?
1) Gathering the data
E (5,1.5) Circumcenter
H (4.3,2.3) incenter
I (3.6, 2.6) is the centroid.
2) Examining the figure we can see point C and B as the vertices of the
triangle, to find the radius let's use the distance formula between point E and C
E(5, 1.5) and C(3,5)
[tex]\begin{gathered} d=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)}^2 \\ \\ d=\sqrt[]{(5-3_{})^2+(1.5_{}-2.6_{})}^2 \\ d=2.28 \end{gathered}[/tex]Since the radius is a line segment from the origin to the circumference then the distance BC = radius of the circumscribed triangle
Radius = 2.28
A bag contains 3 gold marbles, 10 silver marbles, and 23 black marbles. You randomly select one marblefrom the bag. What is the probability that you select a gold marble? Write your answer as a reduced fractionPlgold marble)
ANSWER
P(gold marble) = 1/12
EXPLANATION
In total, there are:
[tex]3+10+23=36[/tex]36 marbles in the bag, where only 3 are gold marbles.
The probability is:
[tex]P(\text{event)}=\frac{\#\text{times the event can happen}}{\#\text{posible outcomes}}[/tex]In this case, the number of posible outcomes is 36, because there are 36 marbles in the bag. The number of times the event can happen is 3, because there are 3 gold marbles:
[tex]P(\text{gold marble)}=\frac{3}{36}=\frac{1}{12}[/tex]26÷2.40=10.833333 round to the nearest cent
26÷ 2.40= 10.833333
Nearest cent means ,2 numbers after decimal point
Then it is 10.83
count 2 numbers to right ,and discard rest of 3333
Then answer is = 10.83
Helen has a box of marbles. 1/2 of the marbles are yellow. 1/8 of the
marbles are red. The rest of the marbles are blue. Helen pulls one marble
out of the box at random, records its color, replaces it, and mixes up the
marbles again. If she does this 400 times, how many blue marbles should
she expect to pull out?
Answer:
150 blue marbles
Step-by-step explanation:
Hello!
If 1/2 of the marbles are yellow, and 1/8 of the marbles are red, then 3/8 of the marbles should be blue.
The percentages are as given:
Yellow = 50%Red = 12.5%Blue = 37.5%To calculate the possible number of blue marbles out of the 400 marbles, we can find 37.5% of 400, as there is a 37.5% chance of getting blue for each turn.
Calculate37.5% of 4000.375 * 400150Helen should expect to pick out 150 blue marbles.
2x + 37 = 7x + 42x = ???
Solve;
[tex]\begin{gathered} 2x+37=7x+42 \\ \text{Collect all like terms and you'll have,} \\ 2x-7x=42-37 \\ \text{Note that a positive number becomes negative once it crosses the equality sign} \\ \text{And vice versa for a negative number} \\ 2x-7x=42-37 \\ -5x=5 \\ \text{Divide both sides by -5} \\ \frac{-5x}{-5}=\frac{5}{-5} \\ x=-1 \end{gathered}[/tex]Therefore, x = -1
= Homework: Module 17If r(x) =find r(a) and write the answer as one fraction.X-29r(a) =(Simplify your answer. Do not factor.)
As given by the question
There are given that function
[tex]r(x)=\frac{7}{x-2}[/tex]Now,
To find the value of r(a^2), put x = a^2 into the function
Then,
[tex]\begin{gathered} r(x)=\frac{7}{x-2} \\ r(a^2)=\frac{7}{a^2-2} \end{gathered}[/tex]Hence, the function is shown below:
[tex]r(a^2)=\frac{7}{a^2-2}[/tex]
Function A and Function B are linear functions.
Which statement is true?
The y-value of Function A when x = -2 is greater than the y-value of Function B when x = -2.
The y-value of Function A when x = -2 is less than the y-value of Function B when x = -2.
Answer:
Step-by-step explanation:
The y-value of Function A when x = - 2 is less than the y-value of Function B when x = - 2.
which example would be likely to give a valid conclusion?
Given: Different statement
To Determine: Which of the statement would give a valid conclusion
Solution
Please note that the statement must be a true representation of the population
Please help me my answer is correct or no
Answer:
the answer is c actully
Step-by-step explanation:
iv'e took that test b4 so you welcome
what is the range of the number of goals scored?
The minimum number of goals scored is 0 and maximum number of goals scored is 7. The range is equal to difference between maximum number of goals and minimum number of goals.
Determine the range for the goals scored.
[tex]\begin{gathered} R=7-0 \\ =7 \end{gathered}[/tex]So answer is 7.
without dividing, how can you tell which quotient is smaller, 30:5 or 30:6 ? eXPLAIN
Without dividing, we can tell that 30:6 has smaller quotient between 30:5 and 30:6.
According to the question,
We have the following two expressions:
30:5 and 30:6
Now, we can easily find which expression has a smaller quotient when the dividend is the same. We need to look at the divisor. If the dividend is the same then the quotient will be smaller for the one with the greater divisor.
In this case, 30:6 has a greater divisor than 30:5 (6 is larger than 5). So, it will have smaller quotient.
Now, we can prove this by dividing both the expressions.
30/6 = 5
(So, it has smaller quotient.)
30/5 = 6
Hence, 30:6 has smaller quotient than 30:5.
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Find the slope of the line that goes through the given points 9,7 and 8,7
we have the points
(9,7) and (8,7)
Note that: The y-coordinates of both points are equal
that means
we have a horizontal line
therefore
The slope is zeroI need to figure out the easiest way to solve this and apply the method to every problem
The function is given as,
[tex]f(x_{)=-3x^2-7x}[/tex]It is asked to find the value of the expression,
[tex]f(7)[/tex]This can be obtained by replacing 'x' by 7 in the given expression of the function,
[tex]f(7)=-3(7)^2-7(7)[/tex]Resolve the parenthesis,
[tex]\begin{gathered} f(7)=-3(49)-49 \\ f(7)=-147-49 \end{gathered}[/tex]Simplify the terms further,
[tex]f(7)=-196[/tex]Thus, the value of the expression f(7) is obtained as,
[tex]=-196[/tex]write your answer in exponential form. 3^9 * 3^-3
Step 1
Given;
[tex]3^9\times3^{-3}[/tex]Required; To write the answer in exponential form
Step 2
[tex]\begin{gathered} Using\text{ the index law below;} \\ a^b\times a^c=a^{bc} \\ Hence,\text{ 3}^9\times3^{-3}=3^{9-3}=3^6 \end{gathered}[/tex]Answer;
[tex]3^6[/tex]Yesterday Ali had n Baseball cards. Today he gave away 6. Using n, Write an expression for the number of cards Ali has left
Yesterday Ali had n Baseball cards.
Today he gave away 6 cards.
We are asked to write an expression for the number of cards Ali has left.
Ali had a total of n cards and he gave away 6 from them.
So, we have to simply subtract 6 cards from the total n cards.
[tex]n-6[/tex]Therefore, the expression is n - 6 represents the number of cards Ali has left.
What is the solution to the equation below?A.x = B.x = C.x = D.x =
Explanation
We are given the following equation:
[tex]\sqrt{5x-2}-1=3[/tex]We are required to determine the value of x.
This is achieved thus:
[tex]\begin{gathered} \sqrt{5x-2}-1=3 \\ \text{ Add 1 to both sides} \\ \sqrt{5x-2}-1+1=3+1 \\ \sqrt{5x-2}=4 \\ \text{ Square both sides } \\ (\sqrt{5x-2})^2=4^2 \\ 5x-2=16 \\ \text{ Collect like terms } \\ 5x=16+2 \\ 5x=18 \\ \text{ Divide both sides by 5} \\ \frac{5x}{5}=\frac{18}{5} \\ x=\frac{18}{5} \end{gathered}[/tex]Hence, the answer is:
[tex]x=\frac{18}{5}[/tex]Which factoring do we use and why and how to know the difference between factoring simple trinomial and perfect square
By definition, a perfect square trinomial is a trinomial that can be written as the square of a binomial. It is in the form:
[tex]a^2+2ab+b^2=(a+b)(a+b)[/tex]The simple trinomial is in the form:
[tex]ax^2+bx+c[/tex]Not all the simple trinomials can be written as the square of a binomial, then we need to check if the trinomial follows the structure of the perfect square trinomial. If it doesn't, then the factors won't be the same, and this is the main difference.
a. The given trinomial is:
[tex]x^2+5x+6[/tex]If it is a perfect square trinomial then:
[tex]\begin{gathered} a^2=x^2 \\ a=x \\ b^2=6 \\ b=\sqrt[]{6} \\ 2ab=5x \\ 2\cdot x\cdot\sqrt[]{6}\ne5x \\ \text{Then it is not a perfect square trinomial} \\ x^2+5x+6=(x+3)(x+2)\text{ It is a simple trinomial} \end{gathered}[/tex]b. The given trinomial is:
[tex]x^2+6x+9[/tex]Let's check if it is a perfect square trinomial:
[tex]\begin{gathered} a^2=x^2\to a=x \\ b^2=9\to b=\sqrt[]{9}=3 \\ 2ab=2\cdot x\cdot3=6x \\ \text{This is a perfect square trinomial, then } \\ x^2+6x+9=(x+3)(x+3)=(x+3)^2 \end{gathered}[/tex]Find two unit vectors orthogonal to both j-k and i+j.
The two unit vectors orthogonal to both j-k and i+j are [tex]\frac{i}{\sqrt{3} }- \frac{j}{\sqrt{3} } -\frac{k}{\sqrt{3} }[/tex]
Let a bar = j- k = < 0,1,-1>
b bar = i+j = <1,1,0>
the cross product a x b bar is orthogonal to both a and b bar
= i ( 0-(-1) ) -j ( 0-(-1) ) + r (0-1)
= i-j-k
A unit vector is a vector whose length is 1 unit
There the unit vector is :
[tex]\frac{i-j-k}{\sqrt{1^2+(-1)^2+(-1)^2} } = \frac{i-j-k}{\sqrt{3} }[/tex]
= [tex]\frac{i}{\sqrt{3} }-\frac{j}{\sqrt{3} }-\frac{k}{\sqrt{3} }[/tex]
The second unit vector orthogonal to both a and b bar would be negative of the previous vector.
= [tex]-\frac{i}{\sqrt{3} }-\frac{j}{\sqrt{3} }-\frac{k}{\sqrt{3} }[/tex]
Hence the two unit vectors orthogonal to both j-k and i+j are [tex]\frac{i}{\sqrt{3} }- \frac{j}{\sqrt{3} } -\frac{k}{\sqrt{3} }[/tex]
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I have tried but but there is some part that i keep getting wrong
we have that
K is the center of circle
J -----> point of tangency
segment IK is a radius
segment JL is a chord
segment GI is a secant
segment JI is a diameter
segment GJ is a tangent
arc JIL is a major arc
arc JL is a minor arc
arc JLI is a half circle (180 degrees)
Part 2
we have that
arc TU=87 degrees -------> by central anglearc ST
Remember that
arc ST+87+72=180 degrees ------> by half circle
so
arc ST=180-159
arc ST=21 degreesarc WV
we have
arc WV+arc UV=180 degrees -----> by half circle
arc UV=72 degrees
so
arc WV=180-72
arc WV=108 degreesarc VUT
arc VUT=arc VU+arc UT
substitute given values
arc VUT=72+87
arc VUT=159 degreesarc WU=180 degrees -----> by half circle deHelp
Show work please
Answer:
check the attached files.