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masya89 [10]
3 years ago
9

How to find a2 and b2 with only c2

Mathematics
1 answer:
Aleks04 [339]3 years ago
5 0

Answer:

The formula is A2 + B2 = C2, this is as simple as one leg of a triangle squared plus another leg of a triangle squared equals the hypotenuse squared.

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Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention
Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

a=\frac{d^2s}{dt^2}

Differentiate the position vector with respect to t.

\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}

Differentiate both sides of the obtained equation with respect to t.

\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}

The initial position is obtained at t=0. Substitute t=0 in the given position function.

\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}

8 0
2 years ago
A class is selling magazines as a fund raiser. Of the 500 magazines sold, Amy sold 1/9 of them. Garret sold 0.145 of the magazin
andrey2020 [161]
500 total magazines...

Amy sold 1/9......1/9(500) = 500/9 = 55.5
Garret sold 0.145.....0.145(500) = 72.5
Robbie sold 12/100.....0.12(500) = 60
Carlita sold 0.095.....0.095(500) = 47.5

Garret sold the most

or u could have done it this way...
Amy sold 1/9.....1/9 = 11.1%
Garret sold 0.145 = 14.5%
Robbie sold 12/100.....12/100 = 12%
Carlita sold 0.095....0.095 = 9.5%

u will still get the same answer...Garret

3 0
3 years ago
Read 2 more answers
An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand
Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

b) We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

Step-by-step explanation:

Previous concepts

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".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

4 0
3 years ago
Will give brainliest
ehidna [41]

Answer:

25.047 or roughly 25.

Step-by-step explanation:

I solved it wrong the first time then i double check with wolframalpha and got the correct number.

We know that the area of the tile is 18 \sqrt{3} and it makes a triangle, we also know the base and height. In this case the base is 2\sqrt{3}^{x/12} and the height is also given which is \sqrt{3}^{\frac{x}{6} }.

Area of triangle = \frac{bh}{2},

substituting we will end up with  18 \sqrt{3}  =( 2\sqrt{3}^{x/12}  *  \sqrt{3}^{\frac{x}{6} }) / 2

Here is the tricky part \sqrt{x} = x^{\frac{1}{2} }, i totally forgot about that lol.

Simplifying: we will get  18 \sqrt{3} = 3^{\frac{x}{8} }

Now, in order to find the x, we will need to take the log of both sides.

log18\sqrt{3}  = \frac{x}{8} log3

Solving for x we end up getting:

8\frac{log18\sqrt{3} }{log3} = x

where x = 25.047.

To be honest I deserve a nobel prize not a brainliest lol.

Good question bro, take it easy.

5 0
3 years ago
What is the y-coordinate of the point that divides the directed
e-lub [12.9K]

Answer:

(D)5

Step-by-step explanation:

Given the point J(-3,1) and K(8,11).

The line segment that divides the segment from J to K in any given ratio can be determined using the formula.

P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n} ,\dfrac{my_2+ny_1}{m+n}\right)

In the given case:

(x_1,y_1)=(-3,1), (x_2,y_2)=(8,11), m:n=2:3

Since we are to determine the y-coordinate of the point that divides JK into a ratio of 2:3, we have:

\dfrac{my_2+ny_1}{m+n}=\dfrac{2*11+3*1}{3+2}\\\\=\dfrac{22+3}{5}\\\\=\dfrac{25}{5}\\\\=5

The y-coordinate of the point that divides the directed  line segment from J to K into a ratio of 2:3 is 5.

The correct option is D.

3 0
3 years ago
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