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3 years ago

Caiden wants to use the distributive property to find the value of 28⋅19+28⋅6528·19+28·65. Which expression can he use?

Mathematics

1 answer:

MakcuM [25]3 years ago

4 0

28 times 28 times (19+65)

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Select ALL of the symbols that would make the comparison true. 2.5 ___ 2.05.

Nutka1998 [239]

I’m pretty sure it would be just E and F

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3 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

Hellppp I'm slowee don't know anything

miss Akunina [59]

Answer: B) X=40

cuz i just know

5 0

3 years ago

A semi regular tesselation may have

AleksandrR [38]

Answer:

Three types of shapes

Step-by-step explanation:

5 0

3 years ago

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The area of a rectangle is represented by x^2- 7x + 10. What are the missing side lengths of the rectangle?

ladessa [460]

Length = x - 2
Width = x - 5

Explanation:

Area = x^2 - 7x + 10
Area = (x - 2)(x - 5)
Area = Length x Width

Length = x - 2
Width = x - 5

4 0

3 years ago