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3 years ago

If f(x) = 2x2 and g(x)= x2 + 2x, which expression represents f(g(x))?

Mathematics

1 answer:

andre [41]3 years ago

7 0

Answer:

f(g(x)) = 2(x^2 + 2x)^2

f(g(x)) = 2x^4 + 8x^3 + 8x^2

Step-by-step explanation:

Given;

f(x) = 2x^2

g(x) = x^2 + 2x

To derive the expression for f(g(x)), we will substitute x in f(x) with g(x).

f(g(x)) = 2(g(x))^2

f(g(x)) = 2(x^2 + 2x)^2

Expanding the equation;

f(g(x)) = 2(x^2 + 2x)(x^2 + 2x)

f(g(x)) = 2(x^4 + 2x^3 + 2x^3 + 4x^2)

f(g(x)) = 2(x^4 + 4x^3 + 4x^2)

f(g(x)) = 2x^4 + 8x^3 + 8x^2

Hope this helps...

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Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$

Now, put the value of length(2B) into equation 3 to know length (1B)

$length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm$

I really hope this helps you, have a great day.

6 0

3 years ago

What is the value of n?<br>enter answer in box

nalin [4]

Answer:

Step-by-step explanation:

chord x chord = chord x chord

(5) (n+8) = (7) (n+4)

i just started plugging in numbers

(5) (6+8) = (7) (6+4)

5 x 14 = 7 x 10

70=70

4 0

3 years ago

Solve the following dividend problem.alameda tent company has 75,000 shares of stock outstanding. what total dividend declaratio

butalik [34]

75,000×2=150,000
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5 0

3 years ago

Read 2 more answers

Need help with number 43 and 45 and need to show the work

Alex787 [66]

43:
   Negative temperatures are basically the same as normal ones, although they go below zero, so they are still the same distance apart from zero, just on the left side of the number line.

      To find the answer find the difference between 30.5°C and 0 (which is 30.5).

      Next, find the difference between -122.2°C and zero (122.2)

   Finally, add 122.2 + 30.5 to find the total distances from the zero combined for the full distances apart from each other.

   (We add them because -122.2 to the left of zero, and 30.5 is on the right)

      You should get 152.7 as your answer.

45:
   I'm guessing since it is asking for the deepest point (-5315) from sea level (0), the answer is -5,315 feet

   *Remember 5,315 is negative because numbers below sea level are all negative*

   So the answer should be -5,315 feet deep. (I'm not sure how to show your work for that I'm guessing it is 0 - 5,315 at the slightest).

*Hope this all helped :)*

5 0

3 years ago

Write the decimal equivalent for this fraction 70/100

d1i1m1o1n [39]

Answer:

0.7 is the decimal equivalent to 70/100.

Step-by-step explanation:

You can just do the numerator (70) divided by the denominator (100): 70÷100 = 0.7

8 0

2 years ago