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3 years ago

Solve for x, check the image below

Mathematics

2 answers:

weeeeeb [17]3 years ago

3 0

Maybe 6 i am not sure

LuckyWell [14K]3 years ago

3 0

X = 8 because the radius is 6. 6 + 4 is 10 so that is your hypotenuse. from there you can just use the pythagorean theorem.

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Look at the figure: An image of a right triangle is shown with an angle labeled x. If tan x° = e divided by 8 and sin x° = e div

Nastasia [14]

Answer:

cos x = 8/f.

Step-by-step explanation:

We can use the identity :

tan x = sin x / cos x.

Substituting:

e/8 = e/f / cos x

e/8 * cos x = e/f

cos x = e/f * 8/e

cos x = 8e / fe

cos x = 8/f.

8 0

3 years ago

Read 2 more answers

One method for estimating abundance of animals is known as line-intercept sampling. The theory of this method, when applied to A

Kruka [31]

Answer:

0.6848

Step-by-step explanation:

Mean of \hat{p} = 0.453

Answer = 0.453

Standard deviation of \hat{p} :

= \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.453(1-0.453)}{100}} = 0.0498

Answer = 0.0498

P(0.0453 - 0.05 < p < 0.0453 + 0.05)

On standardising,

= P(\frac{0.0453-0.05-0.0453}{0.0498} <Z<\frac{0.0453+0.05-0.0453}{0.0498})

= P(-1.0044 < Z < 1.0044) = 0.6848

Answer = 0.6848

3 0

3 years ago

Help please thank you

puteri [66]

What do you need help with? Just lmk

7 0

3 years ago

Read 2 more answers

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago

Am I doing this right!

Leni [432]

16 thru 23 look wrong and 16 is absolutely wrong. You use this:https://www.mathpapa.com/algebra-calculator.html to work out the answers and it shows the steps and explains it to you better than I could. Good luck

4 0

3 years ago