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3 years ago

A chair for an adult is 20 in. tall and 16 in. wide. A builder

Mathematics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

4 in. wide

Step-by-step explanation:

20/5=4

4x=16

4x/4=16/4

x=4

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Compare the fractions. Use <, >, or =.<br> 6<br> 1<br> 7<br> 7

AURORKA [14]

Answer:

-6/7 < -1/7

because it is more than negative one by seven

8 0

2 years ago

What two numbers multiply to -52 and add up to 24

harkovskaia [24]

Answer:

-2 and 26

Step-by-step explanation:

Multiply to -52:

-2 * 26= -52

Add to 24

-2+26= 24

8 0

2 years ago

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Angle C has what measurement according to the protractor?

trasher [3.6K]

The answer is A.50 just take the test. and was write

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3 years ago

The price of a technology stock has dropped to $9.71 today. Yesterday's price was $9.80. Find the percentage decrease.

Vadim26 [7]

Answer:

The percentage decrease = 0.9%

Step-by-step explanation:

Initial value = $9.80

Final value = $9.71

Using the formula

Percentage Decrease = [ (Initial Value - Final Value) / |Initial Value| ] × 100

= [(9.80 - 9.71 ) / (9.80)] × 100

= [0.09 / 9.80] × 100

= 0.009 × 100

= 0.9%

Therefore, the percentage decrease = 0.9%

4 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

3 years ago