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Arturiano [62]
3 years ago
11

How many times larger is 70 than 7

Mathematics
1 answer:
ozzi3 years ago
4 0

the answer is 10! because 7*10= 70

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-2{-7-2[3 squared + 2 to the power of 4]-8÷4}
Leto [7]

Answer:

118 i think

Step-by-step explanation:

-2{-7-2[3 squared + 2 to the power of 4]-8÷4}

-2{-7-2{9+16]-2}

-2{-7-50-2}

-2(-59)

118

5 0
3 years ago
What is the missing side
AveGali [126]

The Pythagorean theorem applies to the sides of this right triangle.

... x² + (4/7)² = (5/7)²

... x² = 25/49 -16/49

... x = √(9/49)

... x = 3/7

The unknown side has length 3/7.

_____

You might recognize that this is a 3-4-5 right triangle with a scale factor of 1/7.

7 0
3 years ago
Read 2 more answers
I am redesigning a rectangular prism in geometry where I have to minimize the surface area while maintaining the same volume. I
Romashka [77]

Strategy:

You will need to make use of the formulas for volume and area of the shapes you have chosen.

... volume of a sphere = (4/3)πr³

... area of a sphere = 4πr²

You will have to use the volume equation to find the radius of the sphere with the desired volume. Then use the area formula to verify the area is smaller than for your rectangular prism. (A sphere has the smallest surface area for a given volume of any figure.)

___

For your pyramid, it's a bit more complicated. You can use the formula for volume to find a relationship between the side length and the height. Then you can solve for height and put that expression into the equation for surface area. Now, you have an equation for surface area as a function of side length, which you can use to choose side lengths that give surface area in the range you want. You may find a graphing utility to be helpful for this.

... volume of a pyramid = (1/3)b^2h . . . . . b is the edge length of the base; h is the verical height.

... area of a pyramid = b^2 + 2bs, where s is the slant height: s=√((b/2)^ +h^2)

___

The cylinder is solved essentially the same way the pyramid is solved, except the formulas are slightly different.

... volume of a cylinder = πr²h

... area of a cylinder = 2πr² + 2πrh

Numbers:

If I did it right, a pyramid with a base edge length of 4 or 5 inches (or somewhere between) should give a smaller surface area for the volume you want.

For the cylinder, the radius may be around 2 or 2.1 inches.

5 0
3 years ago
Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!
LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}

=\sqrt{40}b^2\sqrt{24b^3}

\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

=\sqrt{24}b^{\frac{3}{2}}

=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2

=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

4 0
3 years ago
Can someone help please
Pepsi [2]

Answer:

A: 5

B: 6

C: 3x^2 - 2x

D: 8

E: 22

Step-by-step explanation:

A: (-1)^2 + 4 = 5

B: (-4)^2 +4 = 16

2 x (-3) -4 = -10,

16-10=6

C: 3(x^2 + 4) = 3x^2 + 12

3x^2 + 12 -2x -4, simplify

3x^2 - 2x

D: g(3) = 2*3 -4 = 2

f(2) = 2^2 +4 = 8

E:  f(3) = 3^2 + 4 = 13

g(13) = 2*13 - 4 = 22

6 0
3 years ago
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