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2 years ago

A dog groomer could clean six dogs an hour, how many could they clean five-tents of an hour

Mathematics

1 answer:

Nina [5.8K]2 years ago

5 0

6 times 5 is 30 so 30/10=3 dogs in 5/10 of an hour hooe it helped

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Which pairs of variables would most likely have a positive association? Choose all that apply.

alex41 [277]

Answer:

The age of a kitten and it’s weight

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Step-by-step explanation:

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2 years ago

Find the distance between the pair of points (-7, 2) and (-2, 14)

VladimirAG [237]

Step-by-step explanation:

Distance between points (-7, 2) and (-2, 14) is 13

6 0

3 years ago

What is the answer to this?

Yanka [14]

DEF= 6 ft^2

A= 1/2*b*h

Plug in what you know.

6= 1/2*4*h

Multiply like terms.

6=2*h

Divide 2 on both sides.

3=h

If the base of ABC= 12

12/4=3

3*3=9

A= 1/2*b*h
A= 1/2*12*9
A= 1/2*84
A= 42

The area of ABC is 42 ft^2

I hope this helps!
~kaikers

5 0

2 years ago

I WILL MARK AS BRAINLIEST! Solve for Y

zzz [600]

Answer:

See below.

Step-by-step explanation:

Substitute the value for the variable and solve.

a.)

⇒ y = -5(3) + 17

⇒ y = -15 + 17

⇒ y = 2

b.)

⇒ y = 3(3) - 22

⇒ y = 9 - 22

⇒ y = -13

c.)

⇒ y = 2(3) - 25

⇒ y = 6 - 25

⇒ y = -19

d.)

⇒ y = 6(3) - 39

⇒ y = 18 - 39

⇒ y = -21

7 0

3 years ago

Read 2 more answers

A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.

IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ( $v_0$ ) = 40 ft/sec

Deceleration of the car ( $\frac{dv}{dt}$ ) = -10 ft/sec²

Final speed of the car ( $v_x$ ) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$

Therefore, the car travels a distance of 80 feet before stopping.

4 0

2 years ago