60 is 15% of what number?

For which system of equations is (5, 3) the solution? A. 3x – 2y = 9 3x + 2y = 14 B. x – y = –2 4x – 3y = 11 C. –2x – y = –13 x

Alla [95]

The correct answer is:

D) $\left \{ {{2x-y=7} \atop {2x+7y=31}} \right.$ .

Explanation:

We solve each system to find the correct answer.

For A:
$\left \{ {{3x-2y=9} \atop {3x+2y=14}} \right.$

Since we have the coefficients of both variables the same, we will use elimination to solve this.

Since the coefficients of y are -2 and 2, we can add the equations to solve, since -2+2=0 and cancels the y variable:
$\left \{ {{3x-2y=9} \atop {+(3x+2y=14)}} \right. 
\\
\\6x=23$

Next we divide both sides by 6:
6x/6 = 23/6
x = 23/6

This is not the x-coordinate of the answer we are looking for, so A is not correct.

For B:
$\left \{ {{x-y=-2} \atop {4x-3y=11}} \right.$

For this equation, it will be easier to isolate a variable and use substitution, since the coefficient of both x and y in the first equation is 1:
x-y=-2

Add y to both sides:
x-y+y=-2+y
x=-2+y

We now substitute this in place of x in the second equation:
4x-3y=11
4(-2+y)-3y=11

Using the distributive property, we have:
4(-2)+4(y)-3y=11
-8+4y-3y=11

Combining like terms, we have:
-8+y=11

Add 8 to each side:
-8+y+8=11+8
y=19

This is not the y-coordinate of the answer we're looking for, so B is not correct.

For C:
Since the coefficient of x in the second equation is 1, we will use substitution again.

x+2y=-11

To isolate x, subtract 2y from each side:
x+2y-2y=-11-2y
x=-11-2y

Now substitute this in place of x in the first equation:
-2x-y=-13
-2(-11-2y)-y=-13

Using the distributive property, we have:
-2(-11)-2(-2y)-y=-13
22+4y-y=-13

Combining like terms:
22+3y=-13

Subtract 22 from each side:
22+3y-22=-13-22
3y=-35

Divide both sides by 3:
3y/3 = -35/3
y = -35/3

This is not the y-coordinate of the answer we're looking for, so C is not correct.

For D:
Since the coefficients of x are the same in each equation, we will use elimination. We have 2x in each equation; to eliminate this, we will subtract, since 2x-2x=0:

$\left \{ {{2x-y=7} \atop {-(2x+7y=31)}} \right. 
\\
\\-8y=-24$

Divide both sides by -8:
-8y/-8 = -24/-8
y=3

The y-coordinate is correct; next we check the x-coordinate Substitute the value for y into the first equation:
2x-y=7
2x-3=7

Add 3 to each side:
2x-3+3=7+3
2x=10

Divide each side by 2:
2x/2=10/2
x=5

This gives us the x- and y-coordinate we need, so D is the correct answer.

7 0

3 years ago

The original price of a computer is $489.00. it is discounted 30%. find the selling price

Yanka [14]

Answer:

$342.30

Step-by-step explanation:

The original price of the computer is $489. The computer is discounted 30%, which means that 30% of the price is removed from the original price.

30% of 489 is

489 * 0.3 which is $146.7

146.7 is the discounted amount, not the selling price. To find the selling price, subtract how much is discounted ( 146.7) from the original price (489).

489-146.7= 342.3

3 0

2 years ago

Which equation represents the relationship between the radius, r, and arc length, s?

pochemuha

2pie x r= total arc length of a circle

7 0

3 years ago

Read 2 more answers

Find n in 60/66=n/22

Dmitry_Shevchenko [17]

Answer:

20

Step-by-step explanation:

TO find the answer, use multiplication.

When one part of a fraction multiplies, you multiply the other number with it. 22x3 is 66, so divide 60 by 3. You get 20. n=20. Hope it helps.

You can use ratio. 60:66 is n:22. 22 goes into 66 3 times, and 20 goes into 60 3 times too, so 20:22 is equivalent to 60:66

(please give brainliest)

3 0

2 years ago

Read 2 more answers

Evaluate the infinite sum:

satela [25.4K]

Consider the k-th partial sum,

$S_k = 1 + \dfrac2\pi + \dfrac3{\pi^2} + \cdots + \dfrac k{\pi^{k-1}}$

More compactly,

$\displaystyle S_k = \sum_{i=1}^k \frac i{\pi^{i-1}} = \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}}$

(this is just another case of a similar sum you asked about a while ago [24494877])

The infinite sum is the limit of the partial sum as k goes to infinity. We have

$\displaystyle \lim_{k\to\infty} \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}} = \frac\pi{(1-\pi)^2} \lim_{k\to\infty} \left(\frac{(1-\pi)k}{\pi^k} + \pi - \frac1{\pi^{k-1}} \right) = \boxed{\frac{\pi^2}{(1-\pi)^2}}$

since the non-constant terms in the limit converge to 0.

Alternatively, recall that for |x| < 1, we have

$\dfrac1{1-x} = \displaystyle \sum_{n=0}^\infty x^n$

Differentiating both sides gives

$\dfrac1{(1-x)^2} = \displaystyle \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}$

also valid for |x| < 1. Take x = 1/π and you get the sum you want to compute.

5 0

3 years ago