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3 years ago

What number would you add to both sides of x2 + 7x = 4 to complete the square?

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

Answer: 7/2^2

Step-by-step explanation:

Ed 2020

tamaranim1 [39]3 years ago

5 0

Answer: (7/2)^2

Step-by-step explanation:

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Solve this Equation 10x+2=32

ahrayia [7]

10x+2=32
10x=30
1=3
3=x

7 0

3 years ago

Stefan sells Jin a bicycle for $147 and a helmet for $18. The total cost for Jim is 150% of what Stefan spent originally to buy

soldi70 [24.7K]

Answer:

He spent $110.00 on both originally

Step-by-step explanation:

Add both prices too get 165

Then use the equation $\frac{150}{100}$ = $\frac{165}{x}$

cross multiply to get 150x=16500

Divide 16500 by 150 to get x alone

x= 110

4 0

3 years ago

The statue of liberty is 305 feet tall. A nearby building is 4/9 as tall. Find the height of the building.

inysia [295]

Answer:

The nearby building is $135 \dfrac{5}{9}$ feet tall.

Step-by-step explanation:

The statue of liberty is 305 feet tall.

A nearby building is $\dfrac{4}{9}$ as tall as the statue of liberty.

To find the height of the building, you need to find $\dfrac{4}{9}$ of 305:

$\dfrac{4}{9}\cdot 305=\dfrac{1,220}{9}= 135 \dfrac{5}{9}\ feet$

Thus, the nearby building is $135 \dfrac{5}{9}$ feet tall.

5 0

2 years ago

Write at least 5 arithmetic sequence and determine the common difference

Margarita [4]

Answer: 1,6,11,16,21,26

Step-by-step explanation:

The common difference is 5 (n+5)

7 0

2 years ago

Read 2 more answers

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