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3 years ago

Can you solve #7 if ur not in 8th grade or higher you probably won’t understand but if you do your smart:)

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

7 0

The other leg measures 20.5 m.

Since this is a right triangle, we can use the Pythagorean theorem to solve this. The Pythagorean theorem says that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse, or

a²+b²=c², where a and b are the legs and c is the hypotenuse.

Plugging in the value of the one known leg and the hypotenuse, we have
32²+b²=38²
32*32 + b² = 38*38
1024 + b² = 1444

Subtract 1024 from both sides:
1024 + b² - 1024 = 1444 - 1024
b² = 420

Take the square root of both sides:
√b² = √420
b ≈ 20.5

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What is the value of 4p-2 when p=8

3241004551 [841]

Answer: Hence, the value of expression is 30 when p = 8.

Step-by-step explanation:

Since we have given that

$4p-2$

and we have p = 8.

So, we substitute the value of p in the above expression.

so, it becomes,

$4(8)-2\\\\4\times 8-2\\\\=32-2\\\\=30$

Hence, the value of expression is 30 when p = 8.

3 0

3 years ago

A store sells six cupcakes for $7. How much do 18 cupcakes cost?

ollegr [7]

Answer:

21 cupcakes

Step-by-step explanation:

6 = $7

18 = 6 x 3

18 = 7 x 3

18 = $21

7 0

3 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

1. Solve proportion 12/15 = 18/b

omeli [17]

We solve the as follows:

<span>1. Solve proportion 12/15 = 18/b
</span>( 12/15 = 18/b ) b<span>
( 12b/15 = 18 ) 15/12
b = 18(15/12)
b = 45/2

2. Solve equations

A. -x + 4 = x +6
</span>-x + 4 - x = x +6 - x
<span>-2x + 4 - 4 = 6 - 4
-2x = 2
x = -1

B. 5n + 7 =7(n+1) -2n
5n + 7 = 7n + 7 - 2n
5n -7n + 2n = 7 - 7
0 = 0

C. -4(p+2) + 8 = 2(p-1) - 7p + 15
-4p - 8 + 8 = 2p - 2 - 7p + 15
-4p + 7p - 2p = -2 + 15 + 8 - 8
p =13

3. Solve a/b x - c = 0 for x
</span>a/b x - c = 0
( a/b x = c ) b/a
x = bc / a

3 0

3 years ago

Does anybody know the answer to this question Im having lots of problems

Dvinal [7]

<h3>Answer: 3^9</h3>

This is the same as writing $3^9$

====================================================

How to get that answer:

The lowest height is 3^8, and the highest point is 3 times that value.

We can think of that second "3" as really 3^1. This is because x^1 = x for any real number.

Multiply 3^8 and 3^1 using the rule that a^b*a^c = a^(b+c). We add the exponents together.

Therefore, 3^8*3^1 = 3^(8+1) = 3^9

Side notes:

3^8 = 6,561
3^9 = 19,683

7 0

3 years ago

Read 2 more answers