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2 years ago

Please help me im stuck

Mathematics

1 answer:

Vinvika [58]2 years ago

3 0

Answer: The mean is 6, mean also means the constence.

Step-by-step explanation:

6 is able to go into 10,20,30,40. But I maybe wrong, sorry if I am.

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Scarlett is trying to find the height of a dam. She stands 90 meters away from the dam and records the angle of elevation to the

adoni [48]

Answer:

45.55 meters

Step-by-step explanation:

I'm guessing she recorded above her height, right?

To find the height of the dam, use the tangent ratio:

$tangentX=\frac{opposite}{adjacent}$

The height of the dam is the side opposite the given angle, and this will represent x. The adjacent side (the hypotenuse is NEVER considered the adjacent side) is the distance from Scarlett to the dam. The angle of elevation is the angle of 26°. Insert the values:

$tan26=\frac{x}{90}$

Solve for x. Multiply both sides by 90:

$90*(tan26)=90*(\frac{x}{90})\\\\90*tan26=x$

Insert equation into a calculator:

$x=90*tan26\\\\x=43.9$

Now add Scarlett's height:

$43.9+1.65=45.55$

The dam is 45.55 meters tall.

<em>Finito</em> :D

6 0

4 years ago

Tell whether the ordered pair (-1,3) is a solution of the system of linear equations. y=-7x-4 and y=8x+5

irinina [24]

Answer:

Since it does not plug in to the second linear equation, the ordered pair is not a solution of the linear system.

Step-by-step explanation:

(-1,3); y = -7x -4 y = 8x + 5 3 =7 - 4 3 =/ - 8 + 5

8 0

3 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

4 years ago

LaPeach Elementary School uses 4000 pencils each year. The school has 500 students. What is the unit rate per student?

mezya [45]

Answer:

Step-by-step explanation:

Unit per student = 4000/500 = 8

6 0

3 years ago

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