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dem82 [27]
3 years ago
8

What multiples to -63 and adds to two

Mathematics
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?\
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    276.36 should  be the answer 
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Solve 2/3 X -1/5 >1. X=?
Tresset [83]

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]

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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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8 0
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
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