All categories

4 years ago

The graph of which function has a y-intercept of 3 ?

1 answer:

6 0

It will be whichever equation you see that has the "c" = +3, in the format:
F(x) = y = ax^2 + bx + c, where a, b, and c are all integers, such that plugging a "0" into the x's will give:
y = a•0^2 + b•0 + 3 = 0 + 0 + 3 = 3

You might be interested in

Find the solution set

Marianna [84]

Hi!

Whatever we do to the inequality, we have to do it to both sides.

Subtract 14 from both sides
14 - 14 - 3x < -1 - 14
-3x < -15
Divide by -3 on both sides
-3x/-3 < -15/-3
x < 5
Because we divided by a negative number, the equality sign flips.
x > 5
^ the answer

Hope this helps! :)

4 0

3 years ago

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

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

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

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

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

Subtract 4/9 on both sides:

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

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

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

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

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

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

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

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

4 0

3 years ago

Marlene earned $178 at her summer job. She already had savings of $98.25. She buys a T-shirt for $19.50. How much money does she

Stels [109]

$256.75
Add $178 + $98.25= $276.25
Then $276.25 - $19.50 = $256.75

7 0

3 years ago

Algebra 2, please help! thank you

Hunter-Best [27]

Answer:

-2pi/3

Step-by-step explanation:

y = 2 cos 3(x + 2π∕3) +1

y = A sin(B(x + C)) + D

amplitude is A

period is 2π/B

phase shift is C (positive is to the left)

vertical shift is D

We have a shift to the left of 2 pi /3

7 0

3 years ago

Read 2 more answers

Geometry question plz help

GarryVolchara [31]

Answer:if it’s a pretest just guess you don’t need to get it right

Step-by-step explanation:

5 0

3 years ago