Ask question

Ask question

All categories

3 years ago

12

Please solve for p and q. CALCULUS MATH

1 answer:

frutty [35]3 years ago

4 0

Can someone help me on my question on my page

You might be interested in

Jo had $5 more than Nate and together<br> they had $43. How much did Nate have?

Andru [333]

Answer:

Step-by-step explanation:

43-5

=$38

38/2=19

Nate has $19

hope this helps!

6 0

2 years ago

Read 2 more answers

2(3x + 4) + 6 + 7x<br> What is the answer

Colt1911 [192]

Answer:

13x+14

Step-by-step explanation:

2(3x+4)+6+7x

Distribute:

=(2)(3x)+(2)(4)+6+7x

=6x+8+6+7x

Combine Like Terms:

=6x+8+6+7x

=(6x+7x)+(8+6)

=13x+14

3 0

2 years ago

Read 2 more answers

Which of the following is a polynomial with roots: − square root of 5 , square root of 5 , and −3?

tresset_1 [31]

Roots: - √5 , √5, and - 3

=> these are factors of the polynomial: (x + √5), (x - √5), and (x + 3).

Multiply those three factors:

(x + √5) (x - √5) ( x + 3) = [x^2 - 5] ( x + 3 ) = x^2 + 3x^2 - 5x - 15

Therefore the polynomial x^2 + 3x^2 - 5x - 15 is a polynomial with the given roots.

Answer: option B. x^3 + 3x^2 - 5x - 15

4 0

3 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

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

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

or

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

or

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago

Which two points in the graphed function have an average rate of change of 5?

djverab [1.8K]

Answer:

<h2>D. points A and B</h2>

Step-by-step explanation:

Look at the picture.

$average\ rate=\dfrac{rise}{run}$

$A.\ B\ and\ C\to\dfrac{-3}{-1}=3\\\\B.\ D\ and\ F\to\dfrac{-12}{2}=-6\\\\C.\ C\ and\ E\to\dfrac{-8}{4}=-2\\\\D.\ A\ and\ B\to\dfrac{-5}{-1}=5$

3 0

2 years ago