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3 years ago

Solve: 4|x| = 32 A. 8 B. 24 C. -8 or 8 D. No solution

Mathematics

1 answer:

kirill [66]3 years ago

5 0

<span> C. -8 or 8
</span>
4 |-8| = 4*8 = 32 4 |<span>8</span>| = 4*8 = 32

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If sin²∅ + sin∅ = 1 , then what is the value of sin²∅ + sin⁴∅.

Yakvenalex [24]

Answer:

Step-by-step explanation:

Given sin²∅ + sin∅ = 1, we are to find the value of sin²∅ + sin⁴∅. ... 2

From sin²∅ + sin∅. = 1; sin²∅ = 1 - sin∅. ... 3

Substitute equation 3 into 1

sin²∅ + sin⁴∅

= sin²∅ + (sin²∅)²

= (1 - sin∅)+( 1 - sin∅)²

open the parenthesis

= 1 - sin∅+ (1-2sin∅+ sin²∅)

= 1 - sin∅+ 1-2sin∅+ sin²∅

= 1+1-sin∅-2sin∅+sin²∅

= 2 - 3sin∅+sin²∅

Since sin²∅ = 1 - sin∅, the resulting equation becomes;

= 2 - 3sin∅+(1 - sin∅)

= 2 - 3sin∅+1-sin∅

= 3-4sin∅

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3 years ago

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3 years ago

Solve for a 4=20a a=?

SOVA2 [1]

a would be equal to the number 5

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3 years ago

When is 4x < 4 ? use a drawing to justify your reasoning

pashok25 [27]

4x is less than 4 when x is less than 1

<h3>How to determine when is 4x < 4 ?</h3>

The given statement is

when is 4x < 4

This means that

when is 4x less than 4

So, we have

4x < 4

Divide both sides of the inequality by 4

4x/4 < 4/4

Evaluate the quotient

4x/4 < 1

Evaluate the quotient

x < 1

This means that 4x is less than 4 when x is less than 1

Read more about inequality at:

brainly.com/question/24372553

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2 years ago

Lim 1 - cos 40<br>x>01 - cos 60

Nuetrik [128]

Answer:

The answer is 4/9 if the problem is:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}$ .

Step-by-step explanation:

I think this says:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}$ .

Please correct me if I'm wrong about the problem.

Here are some useful limits we might use:

$\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1$

$\limg_{u \rightarrow 0}\frac{\cos(u)-1}{u}=0$

So for our limit... I'm going to multiply top and bottom by the conjugate of the bottom; that is I'm going to multiply top and bottom by $1+\cos(6\theta)$ :

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{1-\cos(6\theta)}\cdot\frac{1+\cos(6\theta)}{1+\cos(6\theta)}$

When you multiply conjugates you only have to do first and last of FOIL:

$\lim_{\theta \rightarrow 0}\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{1-\cos^2(6\theta)}$

By the Pythagorean Identities, the denominator is equal to $\sin^2(6\theta)$ :

$\lim_{\theta \rightarrow 0}\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{\sin^2(6\theta)}$

I'm going to divide top and bottom by $36\theta^2$ in hopes to use the useful limits I mentioned:

$\lim_{\theta \rightarrow 0}\frac{\frac{(1-\cos(4\theta))(1+\cos(6\theta))}{36\theta^2}}{\frac{\sin^2(6\theta)}{36\theta^2}}$

Let's tweak our useful limits I mentioned so it is more clear what I'm going to do in the following steps:

$\lim_{\theta \rightarrow 0}\frac{\sin(6\theta)}{6\theta}=1$

$\lim_{\theta \rightarrow 0}\frac{\cos(4\theta)-1}{4\theta}=0$

The bottom goes to 1. The limit will go to whatever the top equals if the top limit exists.

So let's look at the top in hopes it goes to a number:

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{36\theta^2} \cdot (1+\cos(6\theta)}$

We are going to multiple the first factor by the conjugate of the top; that is we are multiply top and bottom by $1+\cos(4\theta)$ :

$\lim_{\theta \rightarrow 0}\frac{1-\cos(4\theta)}{36\theta^2} \cdot \frac{1+\cos(4\theta)}{1+\cos(4\theta)} \cdot (1+\cos(6\theta)}$

Recall the thing I said about multiplying conjugates:

$\lim_{\theta \rightarrow 0}\frac{1-\cos^2(4\theta)}{36\theta^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

We are going to apply the Pythagorean Identities here:

$\lim_{\theta \rightarrow 0}\frac{\sin^2(4\theta)}{36\theta^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

$\lim_{\theta \rightarrow 0}\frac{\sin^2(4\theta)}{\frac{9}{4}(4\theta)^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

$\lim_{\theta \rightarrow 0}\frac{4}{9}\frac{\sin^2(4\theta)}{(4\theta)^2} \cdot \frac{1+\cos(6\theta)}{1+\cos(4\theta)}$

Ok this looks good, we are going to apply the useful limits I mentioned along with substitution to find the remaining limits:

$\frac{4}{9}(1)^2 \frac{1+\cos(6(0))}{1+\cos(4(0))}$

$\frac{4}{9}(1)\frac{1+1}{1+1}$

$\frac{4}{9}(1)\frac{2}{2}$

$\frac{4}{9}(1)$

$\frac{4}{9}$

The limit is 4/9.

8 0

3 years ago