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Lera25 [3.4K]
2 years ago
14

Please help me with this question

Mathematics
2 answers:
exis [7]2 years ago
8 0
The answers of your question is 1150
IgorLugansk [536]2 years ago
6 0

Answer:

1150in^3 is close

Step-by-step explanation:

We'll be using the formula for volume of a sphere.

We need the radius - but it's just half the diameter, so r = d/2 = 13in / 2 = 6.5in

V = \frac{4}{3}\pi r^2\\~\\V = \frac{4}{3}\pi \cdot (6.5in)^3 = \frac{4}{3}\pi \cdot 274.625in^3 = \pi \cdot 366.16666...~ in^3 \approx 1150.34650999 in^3\\

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The parent volunteers made 120 snow cones in 5 hours. The student council made 100 snow cones in 4 hours. Who made more snow con
marysya [2.9K]
The parent volunteers:
120 snow cones ... 5 hours
x snow cones = ? ... 1 hour

120 * 1 = 5 * x
120 = 5 * x   /5
x = 120 / 5 = 24 snow cones

The student council:
100 snow cones ... 4 hours
y snow cones = ? ... 1 hour

100 * 1 = 4 * y
100 = 4 * y   /4
y = 100 / 4 = 25 snow cones

Result: The student council made more snow cones per hour (25 > 24).
8 0
3 years ago
Find the next term of the given sequence. 1.31, 2.54, 3.77, ...
ziro4ka [17]
This is an arithmetic sequence with a common difference of 1.23...that means u add 1.23 to each term to find the next term.

1.31 + 1.23 = 2.54
2.54 + 1.23 = 3.77
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7 0
3 years ago
Which triangle congruence is this?<br> SSS<br> SAS<br> ASA<br> AAS<br> HL
MrRissso [65]

Answer:

HL

Step-by-step explanation:

The Hypotenuse-Leg Theorem states that two right triangles are congruent if and only if the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of the other right triangle.

7 0
3 years ago
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4 0
3 years ago
Does this graph have any vertical asymptotes? If so, where?
tresset_1 [31]

Answer:

There are no vertical asympotes for this rational function.

Step-by-step explanation:

For rational functions, a vertical asymptote exists for every value of the independent variable such that function become undefined, that is, such that denominator is zero. Let be the following rational function:

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There is a vertical asymptote for this case:

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Which is out of the interval given to the rational function. Hence, we conclude that there are no vertical asympotes for this rational function.

5 0
2 years ago
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