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vazorg [7]
3 years ago
7

Looking at the two, 7

ula1" title="a^{4} b^{2} c^{2} and14abc^{6}" alt="a^{4} b^{2} c^{2} and14abc^{6}" align="absmiddle" class="latex-formula">

what is the GCF of these
Mathematics
1 answer:
AlexFokin [52]3 years ago
5 0

Answer:

7abc^{2}

Step-by-step explanation:

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52.38 divided by 16 is <span>3.27</span>
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The machinist's goal was to increase his production by at least 10% each day. Assume he achieved his goal. If he was able to mac
koban [17]

Answer:

25+10%= 27.5 Hope this helps.

Step-by-step explanation:


3 0
3 years ago
A cone with radius 5 units is shown below. Its volume is 105 cubic units. Find the height of the cone.
Ainat [17]

Answer:

height of the cone  = 4.01273885 units

Step-by-step explanation:

Volume of a cone(V) is given by:

V = \frac{1}{3} \pi r^2h             ....[1]

where,

r is the radius and h is the height of the cone.

As per the statement:

A cone with radius 5 units and ts volume is 105 cubic units

⇒r = 5 units and V = 105 cubic units

Substitute these given values and use \pi = 3.14 in [1] we have;

105 = \frac{1}{3} \cdot 3.14 \cdot 5^2 \cdot h

Multiply both sides by 3 we have;

⇒315 = 78.5 \cdot h

Divide both sides by 78.5 we have;

4.01273885 = h

or

h = 4.01273885 units

Therefore, the height of the cone is, 4.01273885 units

6 0
4 years ago
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A football player completes 19 of his 25 passes during the season. What percent of his passes did the player complete? Use fract
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19/25 is the answer for this one.
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3 years ago
Evaluate the summation from n equals 1 to infinity of the quotient of 3 and 2 raised the quantity n minus 1 power
Shalnov [3]

Answer:

3*∑ (1/2)^n = 6

Step-by-step explanation:

We have the summation from n = 1 to n = ∞, for:

∑ 3/(2)^(n - 1)

We know that the summation between k = 0, and k = N - 1 for:

∑ r^k = (1 - r^N)/(1 - r)

if we have the summation between k = 0 and k = ∞ - 1 = ∞

(here we used that ∞ is really big, then ∞ - 1 = ∞)

In the numerator we will have the term r^∞, if 0 < r < 1, then r^∞ = 0.

Then if we assume that 0 < r < 1 we can write:

∑ r^k =  1/(1 - r)

In our case, we can rewrite our summation as:

3*∑ (1/2)^n

for n = 0 to  n = ∞

You can see that i changed the limits for n, this does not really matter because we are summing between a number and infinity, the only thing you need to take care is that now the power is n, instead of (n - 1)

Then we have r = (1/2) which is clearly smaller than 1.

then (1/2)

Then this summation is equal to:

3*∑ (1/2)^n = 3*( 1/(1 - 1/2)) = 3*( 1/( 2/2 - 1/2)) = 3*( 1/(1/2)) = 3*2 = 6

3*∑ (1/2)^n = 6

5 0
3 years ago
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