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3 years ago

8

What multiples to -63 and adds to two

1 answer:

ivolga24 [154]3 years ago

4 0

That job can be handled by +9 and -7 working together.

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Mikw earned %ll.76 per hour for working 23.5 hours last week. how much money did mike earn last week?\

ololo11 [35]

276.36 should be the answer

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2 years ago

Solve 2/3 X -1/5 >1. X=?

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Answer:

0 maybe?

Step-by-step explanation:

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2 years ago

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prove that the sum of two consecutive exponents of the number 5 is divisible by 30. If N ia the smaller of two consecutive expon

kotykmax [81]

Answer:

5^n-1

Step-by-step explanation:

5^n-1*30

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2 years ago

The square pyramid pictured below has a surface area of

Evgen [1.6K]

Answer:

Step-by-step explanation:

Triangles

Area of a triangle = 1/2 b * slanted height

b = 6 m

h = 9 m

Area of 1 triangle = 1/2 * 6 * 9

Area of 1 triangle = 27

Area of all 4 triangles = 4*27 = 108 m^2

==================

Base area

The base is a square. All 4 sides are equal.

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Total area

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2 years ago

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Answer this volume based Question. I will make uh brainliest + 50 points

Harlamova29_29 [7]

Answer:

$\huge{\purple {r= 2\times\sqrt[3]3}}$

$\huge 2\times \sqrt [3]3 = 2.88$

Step-by-step explanation:

For solid iron sphere:
radius (r) = 2 cm (Given)

Formula for $V_{sphere}$ is given as:

$V_{sphere} =\frac{4}{3}\pi r^3$

$\implies V_{sphere} =\frac{4}{3}\pi (2)^3$

$\implies V_{sphere} =\frac{32}{3}\pi \:cm^3$

For cone:
r : h = 3 : 4 (Given)
Let r = 3x & h = 4x

Formula for $V_{cone}$ is given as:

$V_{cone} =\frac{1}{3}\pi r^2h$

$\implies V_{cone} =\frac{1}{3}\pi (3x)^2(4x)$

$\implies V_{cone} =\frac{1}{3}\pi (36x^3)$

$\implies V_{cone} =12\pi x^3\: cm^3$

It is given that: iron sphere is melted and recasted in a solid right circular cone of same volume
$\implies V_{cone} = V_{sphere}$

$\implies 12\cancel{\pi} x^3= \frac{32}{3}\cancel{\pi}$

$\implies 12x^3= \frac{32}{3}$

$\implies x^3= \frac{32}{36}$

$\implies x^3= \frac{8}{9}$

$\implies x= \sqrt[3]{\frac{8}{3^2}}$

$\implies x={\frac{2}{ \sqrt[3]{3^2}}}$

$\because r = 3x$

$\implies r=3\times {\frac{2}{ \sqrt[3]{3^2}}}$

$\implies r=3\times 2(3)^{-\frac{2}{3}}$

$\implies r= 2\times (3)^{1-\frac{2}{3}}$

$\implies r= 2\times (3)^{\frac{1}{3}}$

$\implies \huge{\purple {r= 2\times\sqrt[3]3}}$
Assuming log on both sides, we find:

$log r = log (2\times \sqrt [3]3)$

$log r = log (2\times 3^{\frac{1}{3}})$

$log r = log 2+ log 3^{\frac{1}{3}}$

$log r = log 2+ \frac{1}{3}log 3$

$log r = 0.4600704139$

Taking antilog on both sides, we find:

$antilog(log r )= antilog(0.4600704139)$

$\implies r = 2.8844991406$

$\implies \huge \red{r = 2.88\: cm}$

$\implies 2\times \sqrt [3]3 = 2.88$

8 0

2 years ago