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Andrej [43]
3 years ago
10

PLEASE HELP ME BAD AT GEOMETRY PLEASE THANKS A WHOLE BUNCH

Mathematics
1 answer:
Setler [38]3 years ago
5 0
V = π r² h

V = π (4)² 8
 
V = 402.12

Hope this helps :)
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-3x + 1 > 22 or 4x - 5 > 15
Mazyrski [523]

Answer:

Step-by-step explanation:

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2 years ago
Which expressions are equivalent to the one below? Check all that apply. 53 • 5x
weeeeeb [17]
<span>

</span>Calculate 3 x 5, which is 15.Since 15 is two-digit, we carry the first digit 1 to the next column.

Calculate 5 x 5, which is 25. Now add the carry digit of 1, which is 26.Since 26 is two-digit, we carry the first digit 2 to the next column.

Bring down the carry digit of 2.


<span>Therefore, 53 x 5 = 265.</span>

4 0
2 years ago
Read 2 more answers
1) m - 38 = -44 * m = --6 o m = 6 m = 82 O m = --82​
galina1969 [7]

m-38=-44

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hope this helps :)

8 0
3 years ago
The cost of 10 oranges is $1. What is the cost of 5 dozen oranges?
liraira [26]
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5 0
2 years ago
Read 2 more answers
On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

5 0
3 years ago
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