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2 years ago

Terrance ran 5/4 of a mile in 1/4 hour. How far would Terrance run in 1 hour ?

Mathematics

1 answer:

Brrunno [24]2 years ago

7 0

Answer:

Terrance would run 20/4 of a mile or 5 miles

Step-by-step explanation:

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562 + X >= 650
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You flip a coin three times in a row. What is the probability you will get three tails?

Romashka-Z-Leto [24]

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PLEASE LOOK at the photo and help me on these 2 questions :(!!!!!

Tems11 [23]

Answer:

1) f(g(2)) = 24

2) f(g(-1)) = -4

Step-by-step explanation:

1) GIven f(x) = x²+2x and g(x) = 2x

f(g(x)) = f(2x)

f(2x) = (2x)² + 2(2x)

f(2x) = 4x² + 4x

f(g(x)) = 4x² + 4x

f(g(2)) = 4(2)² + 4(2)

f(g(2)) = 16+8

f(g(2)) = 24

2) f(x) = x+1 and g(x) = 5x

f(g(x)) = f(5x)

f(5x)= 5x + 1

f(g(x)) = 5x + 1

f(g(-1)) = 5(-1) + 1

f(g(-1)) = -5+1

f(g(-1)) = -4

8 0

2 years ago

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

2 years ago