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3 years ago

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Please help with problem number 28. It's geometry.

1 answer:

Sindrei [870]3 years ago

3 0

It's geometry, help him with a problem number of 28 please.

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Find the lateral area of the given figure 3cm 3cm 3cm 3cm 3cm

mel-nik [20]

Answer:

$36 cm^2$

Step-by-step explanation:

The Find the lateral area of the given figure = perimeter × height

$\boxed {\mathfrak {given}}:-$

$l=3cm\\\\b=3cm\\\\h= 3cm$

$So, (perimeter) = 2(3+3)$

→ $4$ × $3$ =

→ $12$

<u>lateral area :-</u>

$12$ × $3$

→ $36cm^2$

$----------$

hope it helps..

have a great day!!

4 0

3 years ago

7x-2[2-(6-3x)]<br><br> Can u explain to me how to do this one step by step

aleksandr82 [10.1K]

$7x-2(2-(6-3x))=\\
7x-2(2-6+3x)=\\
7x-4+12-6x=\\
x+8$

3 0

3 years ago

Read 2 more answers

I don’t understand my math homework. Show your work

egoroff_w [7]

Answer: 38 in

Step-by-step explanation:

-There are 4 sides

-Perimeter is the outside

-add 9.5 four times

-9.5+9.5+9.5+9.5=38!

4 0

3 years ago

Read 2 more answers

Find the equation of a circle that has a diameter with the endpoints given by the points A(-4,9) and B (- 2, - 3) )

Aleksandr-060686 [28]

The equation of the circle is $(x+3)^{2}+(y-3)^{2}=37$

Explanation:

Given that the endpoints of the circle are A(-4,9) and B(-2,-3)

We need to determine the equation of the circle.

<u>Center:</u>

The center of the circle can be determined using the midpoint formula,

$Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Substituting the coordinates A(-4,9) and B(-2,-3), we get,

$Center=(\frac{-4-2}{2},\frac{9-3}{2})$

$Center=(\frac{-6}{2},\frac{6}{2})$

$Center=(-3,3)$

Thus, the center of the circle is (-3,3)

<u>Radius:</u>

The radius of the circle can be determined using the distance formula,

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Substituting the center (-3,3) and the endpoint (-4,9), we get,

$r=\sqrt{\left(-4+3\right)^{2}+\left(9-3\right)^{2}}$

$r=\sqrt{\left(-1\right)^{2}+\left(6\right)^{2}}$

$r=\sqrt{1+36}$

$r=\sqrt{37}$

Thus, the radius of the circle is $\sqrt{37}$

<u>Equation of the circle:</u>

The standard form of the equation of the circle is given by

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

where (a,b) is the center and r is the radius.

Substituting the values, we have,

$(x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}$

$(x+3)^{2}+(y-3)^{2}=37$

Thus, the equation of the circle is $(x+3)^{2}+(y-3)^{2}=37$

7 0

3 years ago

Never mind this question. I reorganized it and put it out there to be answered.

alukav5142 [94]

Ok glad that you got an answer

4 0

3 years ago