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alexandr1967 [171]
3 years ago
10

Please help meeeeeee

Mathematics
1 answer:
My name is Ann [436]3 years ago
6 0

Answer:

y = 32.0°

Step-by-step explanation:

Using the sine ratio in the right triangle

siny° = \frac{opposite}{hypotenuse} = \frac{9}{17}, hence

y = sin^{-1} (\frac{9}{17} ) ≈ 32.0°

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3x+4y=24 ,4x+3y=22 simultaneous equations simultaneous equ
Vinvika [58]

Answer:

x=64

y=54

Step-by-step explanation:

3x+4y=24 4(24-4y\3)+3y=22

x=24-4y/3 96-12y\3+3y=22

x=24-4(54)/3

x=-64 96/3-22=3y/3

y=54

4 0
2 years ago
Alan's aunt gave him $95 to spend on clothes at the mall. He bought 5 shirts that cost $6 each and a pair of pants that cost $17
Daniel [21]

Answer:

he has 48 dollars left to spend on clothes

4 0
2 years ago
SOMEONE ANSWER ASAP ILL EVEN CASHAPP YOU A DOLLAR!!!
devlian [24]

Answer:

2/3

Step-by-step explanation:

-2 and 4 are 6 away

4 and 13 are 9 away

Thus it is 6/9 or 2/3

4 0
3 years ago
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The arnold family arrived at the beach at 10:30 a.m. The spent 3 3/4 hours there. what time did they leave the beach?
Dennis_Churaev [7]
2:15 was the time they left the beach
4 0
2 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%20x%20%5E%7B3%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ - x ^{3} } " alt=
noname [10]

I'm guessing you're given the function y(x)=2-x^3, and you're asked to find the inverse function y^{-1}(x). To do this, swap x and y, then solve for y:

x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}

so that the inverse function is

y^{-1}(x)=\sqrt[3]{2-x}

Just to verify:

y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x

y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x

But in case you're actually only interested in computing the square root, first we note that \sqrt x (the real-valued square root) is only defined as long as x\ge0. So \sqrt{-x^3} is defined as long as -x^3\ge0, or x^3\le0, or equivalently x\le0. Under this condition, we could write

\sqrt{-x^3}=\sqrt{-x\times x^2}=\sqrt{-x}\sqrt{x^2}

We can simplify this further, but we have to be careful. Suppose x=-1. Then x^2=(-1)^2=1. But we get the same result if x=1, since x^2=1^2=1. There are two possible values of x that given the same value of x^2, so to capture both of them, we take \sqrt{x^2}=|x|, the absolute value of x. Then

\sqrt{-x^3}=|x|\sqrt{-x}

We can't simplify the square root term further than this.

3 0
3 years ago
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