NEED help with homework

Pleaseee help, Solve for x and y :3

irina1246 [14]

Answer:

x = $\sqrt{6}$ , y = 2 $\sqrt{3}$

Step-by-step explanation:

using the tangent ratio and the exact value tan45° = 1 , then

tan45° = $\frac{opposite}{adjacent}$ = $\frac{x}{\sqrt{6} }$ = 1 , then

x = $\sqrt{6}$

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using the cosine ratio in the right triangle and the exact value

cos45° = $\frac{1}{\sqrt{2} }$ , then

cos45° = $\frac{adjacent}{hypotenuse}$ = $\frac{\sqrt{6} }{y}$ = $\frac{1}{\sqrt{2} }$ ( cross- multiply )

y = $\sqrt{6}$ × $\sqrt{2}$ = $\sqrt{12}$ = 2 $\sqrt{3}$

6 0

2 years ago

Can someone please help

r-ruslan [8.4K]

Answer: 5y^3-7x^5

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Why is the answer A? Plz explain.

emmasim [6.3K]

Answer:

Step-by-step explanation:

the answer cannot be 5. Why? Because the minimum will occur when 12 and 5 point in opposite directions like both are on the x axis pointing to 0 from the east and pointing to zero coming from the west.

12 - 5 = 7

You cannot get a result less than that.

4 0

3 years ago

Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.

Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0

4 years ago

Iq scores are known to be normally distributed. the mean iq score is 100 and the standard deviation is 15. what percent of the p

kobusy [5.1K]

The 100%-Norm.Dist(value,mean,sd,true) percent of the population has an iq over 115

According to the statement

we have to find that the what percent of the population has an iq over 115.

And from given information,

the mean iq score is 100 and the standard deviation is 15.

And

Normal distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

And by use of this formula, we find the percentage of the population has an iq over 115.

Then the result will comes as 100%.

hence, 100%-Norm.Dist(value,mean,sd,true)

So, The 100%-Norm.Dist(value,mean,sd,true) percent of the population has an iq over 115.

Learn more about Normal distribution here

brainly.com/question/4079902

#SPJ4

6 0

2 years ago