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Whitepunk [10]
3 years ago
9

How do I graph the equation -x + 2y = 2

Mathematics
1 answer:
Rama09 [41]3 years ago
4 0

Answer:


Step-by-step explanation:


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-3 1/6 + 2 7/9 + 5 2/3 estimated
adelina 88 [10]

Answer:

5.28 if you round to the nearest hundredth. Drop a best answer and a like if it helped

3 0
3 years ago
-7-2x=7 (x+8) what does X equal ​
BartSMP [9]

Answer:

x=−7

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

8 0
3 years ago
Can some one explain how to do this please?
AleksandrR [38]
N = x/y because  its in the middle
6 0
3 years ago
Deon has a pizza with a diameter
OverLord2011 [107]

The square box is enough to fit the pizza with a diameter of 10 inches inside. Since the area of the square box is more than the area of the pizza, the pizza fits easily in the square box.

<h3>What is the area of the circle and the square?</h3>

The area of the circle is

Ac = πr² = πd²/4 sq. units

Where r is the radius and d is the diameter of the circle.

The area of the square is given by

As = s² sq. units

Where s is the length of the side of a square.

<h3>Calculation:</h3>

It is given that a pizza(in a circular shape) with a diameter d = 10 in is to be placed in a square box of the same length as the diameter of the pizza.

So,

The area of pizza is

Ap = Ac = πd²/4 sq. units

     = π(10)²/4

     = 25π

     = 78.54 sq. in

Then, the area of the square box with the length same as the diameter of the pizza is,

As = d²

    = 10²

    = 100 sq. in

Since the area of the square is more than the area of the pizza (100 sq. inch > 78.54 sq. inch), the pizza easily fits into the square box.

Learn more about the area of a circle here:

brainly.com/question/15673093

#SPJ1

7 0
2 years ago
A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line
Nataliya [291]

Answer:

Volume is 2000\pi\ m^{3}

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d              (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d                  (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

c = \frac{1}{5}

Therefore,

h(y) = \frac{1}{5}y + 5

Now, the Volume of the pool is given by:

V = \int h(y)dA

where

A = r\theta

A = rdr\ d\theta

Thus

V = \int (\frac{1}{5}y + 5)dA

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta

V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta

V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}

V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}

7 0
3 years ago
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