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dedylja [7]
4 years ago
9

What is the solution to the system of equations represented by the two equations?

Mathematics
1 answer:
Svetradugi [14.3K]4 years ago
8 0

Since  (3/2 x)  and  (-1/2 x + 4)  are both equal to the same thing,
they must be equal to each other.

So you can write        3/2 x  =  -1/2 x + 4 .

Use that equation to find the value of 'x'.

Then write that value in place of 'x' in either one
of the original equations to get 'y'.

Now you can solve it yourself, faster than I can write the solution here.

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What is the total surface area of rectangular prism?
algol13

Answer:

162 [m²].

Step-by-step explanation:

required surface is:

A=(3*7+6*7+3*6)*2=162 [m²].

4 0
3 years ago
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Geometry help please
aniked [119]

Answer:

C

Step-by-step explanation:

When cut in half each side is the same as the other

7 0
3 years ago
Michael averages 5 three-pointers in his first 4 basketball games. How many does he need to score at his 5th game in order to br
igor_vitrenko [27]

Answer:

10

Step-by-step explanation:

The easiest way to understand this is not the shortest way.

Give Michael 4 fives and then add an x.

5 + 5 + 5 + 5 + x

Now divide by 5 and equate that to 6

(5 + 5 + 5 + 5 + x)/5 = 6              

Combine the left side

(20 + x)/5 = 6

Multiply by 5

(20 + x) = 6*5

20 + x = 30

Subtract 20 from both sides

x = 30 - 20

x = 10

He needs 10 three - pointers to average 6

4 0
3 years ago
The Cartesian coordinates of a point are given. (a) (−3, 3) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0
irina [24]

Answer:

a) (-3, 3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (3√2, 0.75π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-3√2, 1.75π)

b) (4, 4√3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (8, 0.13π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-8, 1.13π)

Step-by-step explanation:

We know that polar coordinates are related to (x, y) coordinates through

x = r cos θ

y = r sin θ

And r = √[x² + y²]

a) For (-3, 3)

(i) x = -3, y = 3

r = √[x² + y²] = √[(-3)² + (3)²] = √18 = ±3√2

If r > 0, r = 3√2

x = r cos θ

-3 = 3√2 cos θ

cos θ = -3 ÷ 3√2 = -(1/√2)

y = r sin θ

3 = 3√2 sin θ

sin θ = 3 ÷ 3√2 = (1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 0.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (3√2, 0.75π)

(ii) When r < 0, r = -3√2

x = r cos θ

-3 = -3√2 cos θ

cos θ = -3 ÷ -3√2 = (1/√2)

y = r sin θ

3 = -3√2 sin θ

sin θ = 3 ÷ -3√2 = -(1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 1.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (-3√2, 1.75π)

b) For (4, 4√3)

(i) x = 4, y = 4√3

r = √[x² + y²] = √[(4)² + (4√3)²] = √64 = ±8

If r > 0, r = 8

x = r cos θ

4 = 8 cos θ

cos θ = 4 ÷ 8 = 0.50

y = r sin θ

4√3 = 8 sin θ

sin θ = 4√3 ÷ 8 = (√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 0.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (8, 0.13π)

(ii) When r < 0, r = -8

x = r cos θ

4 = -8 cos θ

cos θ = 4 ÷ -8 = -0.50

y = r sin θ

4√3 = -8 sin θ

sin θ = 4√3 ÷ -8 = -(√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 1.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (-8, 1.13π)

Hope this Helps!!!

8 0
3 years ago
Evaluate the following: 3 to the power 2 ÷ (2 + 1). 2 3 4 5
olganol [36]

Answer:

3

Step-by-step explanation:

5 0
3 years ago
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