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2 years ago

15

Given the functions f(x) = 4x and g(x) = 2", which of the

1 answer:

Mumz [18]2 years ago

8 0

hey farquuad

answer is a i think

or c

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Refer to the table below if needed.

elena-14-01-66 [18.8K]

A quadrant is the area that is divided into the x and y axes

The quadrants in which tan $\theta$ and cot $\theta$ are positive are I and III

<h3>How to determine the quadrants</h3>

The tangent of an angle is calculated as:

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

While the cotangent of the angle is calculated as:

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

The above equations mean that:

For the tangent and cotangent of an angle to be positive, then the sine and the cosine of the angle must have the same sign.

In the first quadrant, the sine and the cosine angles are positive.
In the third quadrant, the sine and the cosine angles are negative.

Hence, the quadrants in which tan $\theta$ and cot $\theta$ are positive are I and III

Read more about trigonometry ratios at:

brainly.com/question/8120556

7 0

3 years ago

Can anyone help me on this question please.

LuckyWell [14K]

Answer:

29

Step-by-step explanation:

6 0

3 years ago

Solve the equation 12a3 + 12a2 = -3a. Show all work.

Artist 52 [7]

12a³ + 12a² = -3a | + 3a

12a³ + 12a² + 3a = 0

3a(4a² + 4a + 1) = 0

3a(2a + 1)² = 0

3a = 0 or (2a + 1)² = 0

1) 3a = 0 | : 3

a = 0

2) (2a + 1)² = 0

2a + 1 = 0 | - 1

2a = -1 | : 2

a = -1/2

a ∈ {-1/2, 0}

Used formula:

(a + b)² = a² + 2ab + b²

7 0

3 years ago

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Gala2k [10]

What is the question?

8 0

3 years ago

Surds and roots<br> look at picture

FinnZ [79.3K]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Expand & simplify ⇨ $( \sqrt{10} - \sqrt{2} ) ^{2}$ . Give your answer in the form $b - c \: \sqrt{5}$ where b & c are integers.

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(\sqrt{10}-\sqrt{2}\right)^{2}$ .

$\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}$

The square of $\sqrt{10}$ is 10.

$10-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}$

Factor $10=2\times 5$ . Rewrite the square root of the product $\sqrt{2\times 5}$ as the product of square roots $\sqrt{2}\sqrt{5}$ .

$10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}$

Multiply $\sqrt{2}$ and $\sqrt{2}$ to get 2.

$10-2\times 2\sqrt{5}+\left(\sqrt{2}\right)^{2}$

Multiply -2 and 2 to get -4.

$10-4\sqrt{5}+\left(\sqrt{2}\right)^{2}$

The square of $\sqrt{2}$ is 2.

$10-4\sqrt{5}+2$

Add 10 and 2 to get 12.

$\boxed{ \boxed{\bf\:12-4\sqrt{5} }}$

Here, b & c are integers where $\boxed{ \sf \: b = 12 \: and \: c = 4}$

7 0

3 years ago

Read 2 more answers