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2 years ago

15

Find (fxg)(a)?thanks !!

1 answer:

Ghella [55]2 years ago

5 0

Answer:

12a^2+13a+3

explanation:

multiply:

(3a+1)(4a+3)=(fxg)

12a^2+13a+3

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Find exact values for sin θ, cos θ and tan θ if csc θ = 3/2 and cos θ < 0.

kumpel [21]

Answer:

Part 1) $sin(\theta)=\frac{2}{3}$

Par 2) $cos(\theta)=-\frac{\sqrt{5}}{3}$

Part 3) $tan(\theta)=-\frac{2\sqrt{5}}{5}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we have

$csc(\theta)=\frac{3}{2}$

Remember that

$csc(\theta)=\frac{1}{sin(\theta)}$

therefore

$sin(\theta)=\frac{2}{3}$

step 2

Find the $cos(\theta)$

we know that

$sin^{2}(\theta) +cos^{2}(\theta)=1$

we have

$sin(\theta)=\frac{2}{3}$

substitute

$(\frac{2}{3})^{2} +cos^{2}(\theta)=1$

$\frac{4}{9} +cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{4}{9}$

$cos^{2}(\theta)=\frac{5}{9}$

square root both sides

$cos(\theta)=\pm\frac{\sqrt{5}}{3}$

we have that

$cos(\theta) < 0$ ---> given problem

so

$cos(\theta)=-\frac{\sqrt{5}}{3}$

step 3

Find the $tan(\theta)$

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=\frac{2}{3}$

$cos(\theta)=-\frac{\sqrt{5}}{3}$

substitute

$tan(\theta)=\frac{2}{3}:-\frac{\sqrt{5}}{3}=-\frac{2}{\sqrt{5}}$

Simplify

$tan(\theta)=-\frac{2\sqrt{5}}{5}$

6 0

3 years ago

Robyn opened a bank account to save her birthday money. it was paying 3.75% interest. then the interest rate went up by 0.65%. h

Savatey [412]

Her new interest rate would be 4.40%

8 0

3 years ago

Read 2 more answers

The letters of ALABAMA and LAMB were placed on separate slips of paper and put into two different bags.

labwork [276]

Answer:

45%

Step-by-step explanation:

There are 5 A's out of a total of 11 letters which is 5/11. 5/11 as a decimal is 0.45 which, as a percentage, is 45%

5 0

2 years ago

Find the surface area<br> Will give brainliest to first answer!!!!

user100 [1]

120 I think I’m not sure

7 0

2 years ago

Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$

And so the equation is

$x^2 - \dfrac{5}{4} + \dfrac{3}{8} = 0$

In order to have integer coefficients, you can multiply both sides of the equation by 8:

$8x^2 - 10 + 3 = 0$

5 0

3 years ago