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3 years ago

I NEED HELP PLEASE ASAP!!

Mathematics

1 answer:

hammer [34]3 years ago

7 0

Answer:

Option B, 1

Step-by-step explanation:

tan 45° = 1/1 = 1

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Simplify 7 - 2(5 - 2x)

igomit [66]

Answer:

4 − 3

Step-by-step explanation:

3 0

3 years ago

List all the pairs of integers with a product of 36. then find the pair whose sum is 20

RoseWind [281]

This is an excellent practice for the solution of quadratic equations.

1*36=36 => (1,36)
2*18=36 => (2,18)
3*12=36 => (3,12)
4*9=36 => (4,9)
6*6=36 => (6,6)
9*4=36 => (9,4)
12*3=36 => (12,3)
18*2=36 => (18,2)
36*1=36 => (36,1)

We can see that the sum decreases until the two factors are close (or equal) and then increases again.

The pair of integers with a sum of 20 is therefore (2,18) or (18,2).

4 0

4 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

I have to pack 296 apples into 8 boxes. How many apples will there be in a box if each box contains the same number of apples

kirill115 [55]

Answer:

Step-by-step explanation:

296÷8=37

Hope this helps! :)

8 0

3 years ago

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Help with 5??????????

crimeas [40]

Answer is number one. The number inside the bracket is the time and the number after the equal sign is the height. As you can see, at t=5, height=75.

8 0

3 years ago

Read 2 more answers