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3 years ago

A number increased by 9 is 13

Mathematics

1 answer:

Alborosie3 years ago

6 0

Answer:

Step-by-step explanation:

To find the number you have to reverse the equation

Let us call the number "x"

x + 9 = 13

So to find x we flip the equation

13 - 9 = x

Solve it

4 = x

So the missing number is 4

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Solve 5x(x-4) using distributive property

vodomira [7]

Answer:

5x² - 20x

Step-by-step explanation:

multiply each term in the parenthesis by the 5x outside

5x(x - 4)

= (5x × x ) + (5x × - 4 )

= 5x² + (- 20x) = 5x² - 20x

4 0

3 years ago

Read 2 more answers

I need help please help

Bumek [7]

Answer:

:) good day mate

Step-by-step explanation:

:) indeed

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4 years ago

What is the X values such that y equals 15 - 3 X and Y equals 0 question

Oksana_A [137]

X=5. this is true because if y=0 and y=15-3x, then 15-3x=0. if x is 5, it would be 15-3(5)=0.

4 0

3 years ago

Two trucks were driven on a 1,680 kilometer (km) trip. the first truck averaged 14 km per liter of fuel for the trip, and the se

kiruha [24]

for the first truck 1680 / 14 = 120 liters used

the second 1680 / 12 = 140 litres used

so 140-120 = 20 liters <span>the second truck used more </span>

8 0

3 years ago

A force of 7 lb is required to hold a spring stretched 9 in. beyond its natural length. How much work W is done in stretching it

Ymorist [56]

144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Force F=10 lb

From the hookes Law

$F=kx$

Therefore, calculate k for the spring for

$10=k(\frac{4}{12}) = > k=\frac{120}{4}=30$

Work done in stretching a spring through a length dx is

$dW=F.dx = > dW=kxdx$

For calculating the work done for stretching the spring to x=6in=(6/12)feet beyond the natural length, Integrate over the limits of x=0 to x=1/2

Therefore,

$\int_{0}^{W}dW=\int_{0}^{0.5}kxdx = > W=\frac{1}{2}k[x^2]_{0}^{0.5}=\frac{1}{2}30 \times (0.5)^2=\frac{30}{8}$

That is,

$W=\frac{15}{4} ft-lb\\$

The answer that is given is for stretching the spring to 4 inches.

Mass of 10 m of chain length is 80kg. This implies, Mass per unit meter length of the chain is

$m_{l}=\frac{80}{10}=8kg/m$

Consider a small length dx of the chain at the end point A. Work done in lifting the small length dx over the height of x meters is

$dW=(8g)dx \times x=8gxdx$

Now, integrate over the the value of x from

x=0 when end of the chain A is on the ground

x=6 when end of the chain A reaches 6 m above the ground

That is,

$\int_{0}^{W}dW=\int_{0}^{6}8gxdx=8g [\frac{x^2}{2}]_{0}^{6}=4g(6^2-0^2)$

$W=4g \times 36=144g \ Joules$

Hence,144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Learn more about Integration here brainly.com/question/2263647

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7 0

2 years ago