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3 years ago

Help !!! 50 points for the right answer!!!!

Mathematics

2 answers:

mart [117]3 years ago

5 0

It doesn't give any answers? I think the first one is 14 the second one is -4 and the third one is 10

kap26 [50]3 years ago

3 0

$A=(14,-6)\qquad B=(-4,7)\\\\\\\bold{u}=\overrightarrow{AB} =\left\ \textless \ -4-14,7-(-6)\right\ \textgreater \ =\left\ \textless \ -18,13\right\ \textgreater \ \\\\\\
||\bold{u}||=\sqrt{(-18)^2+13^2}=\sqrt{324+169}=\sqrt{493}\approx22,2$

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What is equivalent to 5 squared times 9 squared

Pepsi [2]

Answer:

2,025

Step-by-step explanation:

5 squared is 5x5 = 25

9 squared is 9x9 = 81

25x81 = 2,025

4 0

3 years ago

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

What is the distance around a triangle that has sides measuring 2/18 feet 3/12 feet and 2/12 feet

Minchanka [31]

The distance around an object is called the perimeter. To get the perimeter, you add up the sides. So, 2/18 + 3/12 + 2/12 = your answer.

4 0

3 years ago

FInd the value of x. Y D W X (15x-8) X = N 226

Oksi-84 [34.3K]

Let's start with the given arc and its angle.

The angle YWX is going to be half the arc length.

YWX = 1/2 (226) = 113

Angles VWX and YWX form a linear pair (or are supplementary angles), which means that their sum is 180 degrees.

(15x - 8) + 113 = 180

15x + 105 = 180

15x = 75

x = 5

Hope this helps!

8 0

3 years ago

XPF is congruent to which correctly named triangle

guapka [62]

Answer:

the answer is triangle GBL

7 0

4 years ago