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2 years ago

Use the drawing tools to form the correct answer on the graph.

Mathematics

2 answers:

-Dominant- [34]2 years ago

4 0

Answer:

dad

Step-by-step explanation:

dad

CaHeK987 [17]2 years ago

4 0

Answer:

The correct solution is f + goh = x² + 6x +1

Step-by-step explanation:

f (x) = x^2

g(x) = 3x + 1

h(x) = 2x

We are to look for f + (goh)

First we need to get the composite function goh

goh = g(h(x))

g(h(x)) = g(2x)

To get g(2x) we will substitute x in g(x) as 2x as shown below:

Given g(x) = 3x+1

g(2x) = 3(2x)+1

g(2x) = 6x+1

f + goh = x² + (6x +1)

f + goh = x² + 6x +1

Jim wrote goh as 2(3x+1) instead of 3(2x) + 1. The composite function that Jim looked for was hog not goh.

The correct solution is f + goh = x² + 6x +1

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3 years ago

How much simple interest would $1,500 earn in 11 months at an interest rate of 3.75%

FinnZ [79.3K]

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3 years ago

Read 2 more answers

Theres 30 kilograms of 10% apple juice. How many kilograms of apples does it need to increase the current apple juice to 25%?

raketka [301]

The mass of the apple juice needed to increase the initial concentration to 25% is 45 kg

The given parameters;

Mass of the apple juice, m = 30 kg
current percentage, = 10%
final percent after increment = 25%

Let the mass added to increase the percentage to 30% = x

10% ----------------- 30 kg

25% ------------------ ?

$= \frac{25 \times 30}{10} \\\\= 75 \ kg$

The additional mass is calculated as;

$x+ 30 = 75\\\\x = 75 - 30\\\\x = 45 \ kg$

Thus, the mass of the apple juice needed to increase the initial concentration to 25% is 45 kg.

Learn more about percentage and ratio here: brainly.com/question/11752721

6 0

2 years ago

Use series to verify that <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

3 years ago

Chloe is packing up her room to move to a new city. Together, all the books on her bookshelf weigh 85 pounds. She wants to divid

agasfer [191]

She will need 25 pounds 4 ounces in each box

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2 years ago