!!PLZ HELP WILL MARK BRAINLIEST !!

Mathematics

2 answers:

Minchanka [31]1 year ago

7 0

Answer:

2y I think ( I am not sure tho)

KATRIN_1 [288]1 year ago

3 0

The graph is increasing in two intervals

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If 0 < x < 1 and 0 < y < 1, which of the following must be true?

katovenus [111]

0<x<1 and 0<y<1

x>0 so x is positive and y>0 so y is also positive.
When you multiply two positive numbers you always get a positive number, so the product of x and y must be positive, or greater than 0.
xy>0 - it must be true
xy<0 - it can't be true
Also when you divide a positive number by a positive number you always get a positive number, so the quotient of x and y must be positive.
x/y<0 - it can't be true
D and E can be true, but don't have to. It depends on the values of x and y. If x>y, then x-y>0 is true and x-y<0 isn't true; if x<y, then x-y>0 isn't true and x-y<0 is true.

Therefore, only A must be true.

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3 years ago

Help please n thank you

densk [106]

Answer:

Step-by-step explanation:

The non-fillee dots stop at the 98 row and the 0.2 column.

98 + 0.2 = 98.2

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2 years ago

How to write out 314,207

vodomira [7]

Three hundred fourteen thousand, two hundred seven. hope this helps

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2 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: