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Fiesta28 [93]
3 years ago
9

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 &lt;, &gt;, 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
5 0
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
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