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3 years ago

PLEASE HELP ASAP!!

Mathematics

1 answer:

natka813 [3]3 years ago

3 0

Answer:

Since this graph is continuing, y must be any real number. So, d.

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Holly drew the parallelogram below to represent the design of her new garden. A parallelogram with base b and height h. She foun

BabaBlast [244]

Answer: i just took the quiz it is 15

Step-by-step explanation:

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3 years ago

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Express the interval in set-builder notation and graph the interval on a number line (-6.2]

zzz [600]

Answer:

{ $x|-6$ }

The number line is attached.

Step-by-step explanation:

By definition Set-builder notation is a mathematical shorthand used to provide specific details about a set.

In this case, you have the following interval:

$(-6,2]$

You can obsseve that the left hand contains a parentheses, which means that -6 it is not included, then:

$-6$

The right hand contains a bracket, which means that 2 is included:

$\leq 2$

The inequality is:

$-6$

Therefore, in Set-builder notation this is:

{ $x|-6$ }

Observe the number line attached. Since -6 is not included, you must use an unfilled circle. Since 2 is included, you must use a filled circle.

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3 years ago

Is 1/8 greater than 2/10

emmasim [6.3K]

No, 1/8 is smaller because 2/10 is equivalent to 1/5 and 1/5 is larger than 1/8.

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3 years ago

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Suppose cos(x)= -1/3, where π/2 ≤ x ≤ π. What is the value of tan(2x). EDGE

AVprozaik [17]

Answer:

Step-by-step explanation:

We are given that:

$\displaystyle \cos x = -\frac{1}{3}\text{ where } \pi /2 \leq x \leq \pi$

And we want to find the value of tan(2x).

Note that since x is between π/2 and π, it is in QII.

In QII, cosine and tangent are negative and only sine is positive.

We can rewrite our expression as:

$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}$

Using double angle identities:

$\displaystyle \tan(2x)=\frac{2\sin x\cos x}{\cos^2 x-\sin^2 x}$

Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(x) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:

$\displaystyle o=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$

So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.

From the above information, substitute in appropriate values. And since x is in QII, cosine and tangent will be negative while sine will be positive. Hence: