Ask question

Ask question

All categories

2 years ago

7

Does anyone know this answer ?

2 answers:

Arturiano [62]2 years ago

4 0

Answer:

10z + 5

Step-by-step explanation:

NeTakaya2 years ago

4 0

If I’m not mistaken I think is the first one

10z + 5

You might be interested in

JUST HELP ME By what percent will the product of two numbers

Salsk061 [2.6K]

Answer:

The product of the original numbers: xy

The product of the altered numbers (.75x)(.5y), or 0.375xy

The product has decreased by 1 - 0.375, or 0.625, or 62.5%

6 0

2 years ago

Write 83 as a sum of tens and ones

Travka [436]

8 tens and 3 ones.

hope that helped

6 0

2 years ago

Read 2 more answers

What is 6.213 as a mixed number

Ira Lisetskai [31]

6.213 as a mixed number would be 6 and 213/1000.

all you do to convert it to a mixed number is look at how many numbers are after the decimal.

since theres 3 numbers, that means it would be in the thousands place

3 0

3 years ago

Find each key feature of the function shown in the graph. Write the range and domain in interval notation, and assume all values

gayaneshka [121]

The domain of the graph is $-8$ while the range of the graph is $-4\leq x$ .

The domain of the graph is input values of "x" for which the function exists while the range is the output values "y" for which the function exists.

For the graph, the domain will be the values of the graph along the x-axis.

Domain = $-8$ (Note that -8 is not included)

For the range, the interval is given as: $-4\leq x$ . Note that the value of 4 is not included since it is opened.

Learn more on domain and range here: brainly.com/question/1942755

3 0

2 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

6 0

2 years ago