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BARSIC [14]
3 years ago
12

The price of a television was reduced from 250 to 200.By what percentage was the price of the television reduced? A.20 B.25 C.80

D.50 Show work how u got there.
Mathematics
1 answer:
qaws [65]3 years ago
7 0
Hey mate. This is the answer
the change in price was 50 dollars
Now divide your change in price by the original price
50/250=.2
To get that into a percentage multiply it by 100 and you get 20%
The answer is option a

Plssssss mark as brainliest
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Step-by-step explanation:

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Someone help and please answer it fully
IRINA_888 [86]

Answer:

No, Matt did not solve the equation correctly

Correct Answer: x = 8

Step-by-step explanation:

4(x + 2) = 30

Step 1: Distribute

4x + 2 = 30      =>     This is his mistake, he should completely distribute 4    

                                  to x and 2

Step 2: Subtract 2 from both sides/Isolate x

4x = 28           =>      This part is done correctly, but wrong because of Step 1

Step 3: Divide both sides by 4

x = 7                =>      This is correct, but again, he messed up on Step 1

<h3>Let's find the correct answer to this equation:</h3><h3>4(x +2) = 30</h3>

Step 1: Distribute

Remember to distribute 4 to all terms in the parenthesis.

4(x + 2) = 4(x) + 4(2)

            = 4x + 8

4x + 8 = 30

Step 2: Subtract 8 from both sides/Isolate x

Move all the terms that do not belong to x to the other side. We can do this by subtracting 8 from both sides  => (opposite operation of adding 8)

4x + 8 = 30

4x = 30 - 8

4x = 32

Step 3: Divide both sides by 4/Isolate x

Now we want x by itself. Since x is being multiplied by 4, we have to use the opposite operation, dividing by 4, to have x on one side by itself

4x = 32

4(x) = 32

x = 32 ÷ 4

x = 8

-Chetan K

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3 years ago
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skad [1K]

Parameterize S by the vector function

\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

\vec n = \dfrac{\partial\vec s}{\partial v} \times \dfrac{\partial \vec s}{\partial u} = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle

The integral of the field over S is then

\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv

\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv

\displaystyle = 128 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \cos^2(u) \sin^3(v) \, du \, dv = \boxed{\frac{64\pi}3}

8 0
2 years ago
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