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3 years ago

Is Erik's statement correct? Why or why not?

Mathematics

2 answers:

Dafna11 [192]3 years ago

7 0

Answer:

Step-by-step explanation:

givi [52]3 years ago

5 0

Answer:

Step-by-step explanation:

No, the heights of the cones are not the same, so Cavalieri’s principle does not apply.

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Factor the expression using the GCF. The expression 21m−49n factored using the GCF is...

Roman55 [17]

Answer: 7(3m - 7n)

Step-by-step explanation:

Factor each term then see what value(s) they have in common:

21m - 49n

∧ ∧

3 · 7 · m 7 · 7 · n

Both terms have a 7 in common. What is left over after removing the 7?

21m: 7(3m)

49n: 7(7n)

GCF: 7 Leftovers: 3m and 7n

Factored form of expression is: 7(3m - 7n)

6 0

3 years ago

Read 2 more answers

What is the answer to the fraction equation: 2/3+5/6÷1/3?

mash [69]

Answer:

3 1/6

Step-by-step explanation:

8 0

1 year ago

The sphere below has a radius of 2.5 inches and an approximate volume of 65.42 cubic inches.

Stells [14]

Part a: The radius of the second sphere is 5 inches.

Part b: The volume of the second sphere is 523.33 in³

Part c; The radius of the third sphere is 1.875 inches.

Part d: The volume of the third sphere is 27.59 in³

Explanation:

Given that the radius of the sphere is 2.5 inches.

Part a: We need to determine the radius of the second sphere.

Given that the second sphere has twice the radius of the given sphere.

Radius of the second sphere = 2 × 2.5 = 5 inches

Thus, the radius of the second sphere is 5 inches.

Part b: we need to determine the volume of the second sphere.

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=5$ , we get,

$V=\frac{4}{3} (3.14)(125)$

$V=\frac{1580}{3}$

$V=523.3333 \ in^3$

Rounding off to two decimal places, we have,

$V=523.33 \ in^3$

Thus, the volume of the second sphere is 523.33 in³

Part c: We need to determine the radius of the third sphere

Given that the third sphere has a diameter that is three-fourths of the diameter of the given sphere.

Hence, we have,

Diameter of the third sphere = $\frac{3}{4} (5)=3.75$

Radius of the third sphere = $\frac{3.75}{2} =1.875$

Thus, the radius of the third sphere is 1.875 inches

Part d: We need to determine the volume of the third sphere

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=1.875$ , we get,

$V=\frac{4}{3} (3.14)(1.875)^3$

$V=\frac{4}{3} (3.14)(6.59)$

$V=27.5901 \ in^3$

Rounding off to two decimal places, we have,

$V=27.59 \ in^3$

Thus, the volume of the third sphere is 27.59 in³

4 0

3 years ago

The probability of an event happening is (11/30), what is the probability of the event not happening?

Andrew [12]

Answer:

19/30

Step-by-step explanation:

If the probability of an event happening is 11/30, then you simply subtract 11 from 30 to find the remaining "possibilities"

30 - 11 = 19

So- this means that the probability of the event NOT happening, is...

19/30

Hope this helps!

7 0

3 years ago

Multiply or divide as indicated. Leave your answer with no factors in the denominator. p⁻⁴q⁵r⁶/p⁻³qr⁻²

KIM [24]

Answer: $\frac{q^4r^8}{p}$

Step-by-step explanation:

By the Negative exponent rule, we know that:

$a^{-1}=\frac{1}{a}$

By the Quotient of powers property, we know that:

$\frac{a^m}{a^n}=a^{(m-n)$

And by the Product of powers property, we know that:

$(a^m)(a^n)=a^{(m+n)}$

Applying this properties, we get:

$\frac{p^{-4}q^5r^6}{p^{-3}qr^{-2}}=\frac{p^3q^5r^6r^2}{p^4q}=\frac{q^4r^8}{p}$

5 0

3 years ago

Read 2 more answers