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wlad13 [49]
3 years ago
14

Hey,Can someone help me Solve this Math problem ​

Mathematics
2 answers:
Montano1993 [528]3 years ago
6 0

Answer:

120

Step-by-step explanation:

V=1/3Abh

4*10=40

40*9=360

360/3=120

Salsk061 [2.6K]3 years ago
4 0

Answer:

120

Step-by-step explanation:

      l*w*h

v =  --------

         3

l = 10

w = 4

h = 9

so 10 * 4 * 9 = 360

360/3 = 120

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nasty-shy [4]

Answer:

A

Step-by-step explanation:

Adding a negative is the same as subtracting a positive

Keep Change Change

ex. -9.2 - 6.7 = -9.2 + -6.7

    Keep    Change     Change

   -9.2           +              -6.7

6 0
3 years ago
Three students are arguing over who can run the fastest. Student A says that she can run 2/3 mile in six minutes. Student B says
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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
Pls help<br> First, correct answer will get brainliest
kozerog [31]

Answer:

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Step-by-step explanation:

7 0
3 years ago
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KATRIN_1 [288]

Answer:

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4 0
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