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3 years ago

Find The Surface Area. Geometry

Mathematics

1 answer:

Alexeev081 [22]3 years ago

7 0

The pyramid has 4 faces. Each face is a triangle with a base of 12 and a height of 7. The area of any triangle is half of the product (base)x(height). If you also want to include the area of the pyramid's base, that's a square with sides of 12, and the area of a square is the square of the length of a side.

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How do you solve -28-6x=-2(1-3x)-2

Zina [86]

-28-6x=-2(1-3x)-2

-28-6x=-2+6x-2

-28-6x=-4+6x

-6x-6x=-4+28

-12x=24

x=-2

6 0

3 years ago

Read 2 more answers

Plz..... Solve this questions for me...

faust18 [17]

$6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================$

$7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================$

$8.\\5^a\times5^b=5^{11}\\\\a)\\\text{use}\ a^n\times a^m=a^{n+m}\\\\5^a\times5^b=5^{a+b}\\\\5^{a+b}=5^{11}\Rightarrow a+b=11\\\\11\ \text{is odd. The sum of even and odd is odd. Therefore}\ a\ \text{and}\ b\ \text{can't be even.}\\--------------------\\\\b)\\5\ \text{solutions}\\\\11=5+6\\11=4+7\\11=3+8\\11=2+9\\11=1+10$

3 0

2 years ago

A circle is circumscribed around a square and another circle is inscribed in the square. If the area of the square is 9 in2, wha

lesantik [10]

Answer:

√2:1

Step-by-step explanation:

First we need to know that the length of the side of the square is equal to the diameter of the inscribed circle i.e

L = di

Given the area of the square to be 9in², we can get the length of the square.

Area of a square = L²

L is the length of the square.

9 = L²

L = √9

L = 3in

Hence the length of one side of the square is 3in

This means that the diameter of the inscribed circle di is also 3in.

Circumference of a circle = π×diameter of the circle(di)

Circumference of inscribed circle = π×3

= 3π in

For the circumscribed circumscribed circle, diameter of the outer circle will be equivalent to the diagonal of the square.

To get the diagonal d0, we will apply the Pythagoras theorem.

d0² = L²+L²

d0² = 3²+3²

d0² = 9+9

d0² = 18

d0 = √18

d0 = √9×√2

d0 = 3√2 in

Hence the diameter of the circumscribed circle (d0) is 3√2 in

Circumference of the circumscribed circle = πd0

= π(3√2)

= 3√2 π in

Hence, ratio of the circumference of the circumscribed circle to the one of the inscribed will be 3√2 π/3π = √2:1

8 0

3 years ago

a trapezoid has parallel sides of length 26 mm and 44 mm. its other sides measure 10 mm and 17 mm. what is the length of the bim

kirill115 [55]

Answer:

35 mm.

Step-by-step explanation:

That would be (26 + 44)/2

= 70/2

= 35 mm.

5 0

3 years ago

What is the product? Two-thirds times eight-ninths StartFraction 10 over 27 EndFraction StartFraction 16 over 27 EndFraction Two

melomori [17]

Answer:

16/27

Step-by-step explanation:

First let's convert this info an equation and then we will solve.

(2/3) * (8/9) = ?

( 2 * 8 ) / ( 3 * 9 ) = ?

( 16 ) / ( 27 ) = ?

16 / 27 = ?

So our fractional answer is 16/27.

In words this would be sixteen over twenty seven.

Cheers.

3 0

2 years ago