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3 years ago

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There are 4.5 litres of milk in a pot. The chef stirs in 2.5 g of salt and 1.2 litres of water. How much liquid is in the pot?

1 answer:

wlad13 [49]3 years ago

8 0

Answer:

5.7 litres

Step-by-step explanation:

The salt is not a liquid so you do not need to add it in.

4.5 litres + 1.2 litres = 5.7 litres

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what is the area of a regular pentagon with sides of length 13 and an apothem of length 8.95, in square units

Assoli18 [71]

The area is 290.760680699535.

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3 years ago

What’s the equation?

Sphinxa [80]

If Whitney works 51 hours per week, then that would mean that we’d use the expression 51w because w, representing the weeks, would be multiplied by 51 to find the amount of hours worked in that week. However, since 15 of the hours were spent outside instead of in the office, we’d have to subtract that from the 51w. Therefore, the equation would be:

h= 51w - 15

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3 years ago

Read 2 more answers

Write an equation in point slope form for the line with a slope of 3 passes through (1,2)

alexgriva [62]

Answer:

$y-2=3(x-1)$

Step-by-step explanation:

The point-slope form of a line is given as:

$y-y_1=m(x-x_1)$

Where

m is the slope

$x_1$ is the x-coordinate of the point given (passing through the line)

$y_1$ is the y-coordinate of the point given (passing through the line)

Now,

Given,

Slope = m = 3

$x_1$ = 1

$y_1$ = 2

Now, we simply plug these into the formula for point-slope form of a line:

$y-y_1=m(x-x_1)\\y-2=3(x-1)$

This is the point-slope form.

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3 years ago

The braking distance, in feet of a car a Travling at v miles per hour is given.

irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

The braking distance at 25 mph, is approximately <u>63.7 ft.</u>

The braking distance at 55 mph, is approximately <u>298.35 ft.</u>

The braking distance at 85 mph, is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

$D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.$

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945 ft.s

$D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.$

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

$D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.$

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

$450 = 2.6 + \dfrac{v^2}{22}$

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

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2 years ago

What is the midpoint of Line Segment AB, when A(7, 3) and B(7, -7)?

Hatshy [7]

Answer:

(7, -2)

Step-by-step explanation:

Add the points together and divide them by 2:

A+B/2

= ([7 + 7]/2, [3 + -7]/2)

= ([14]/2, [-4]/2)

= (7, -2)

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3 years ago