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3 years ago

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Tracey paid $165 for an item that was originally priced at $570. What percent of the original price did Tracey pay? Round to the

nearest tenth of a percent. percent

1 answer:

kirza4 [7]3 years ago

7 0

Answer:

<h2>28.9%</h2>

Step-by-step explanation:

$\frac{165}{570}$ = 0.2894

multiply by 100 to turn into percentage

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The class size for 8 classrooms is shown. 14, 22, 24, 25, 25, 28, 32, 34 Which class size affects the range of all 8 classes the

dexar [7]

Answer:

B.25

Step-by-step explanation:

I think it's because its the median and i don't think its 14 34 and what does 22 have to do with this.

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2 years ago

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The graph of a quadratic function touches, but does not cross, the x-axis at x = 4. Which function represents this situation?

jek_recluse [69]

When a zero of a function does not cross the x-axis but touches it, this means that the zero is a "double root". It has two of the same factor (x - 4)^2 and two of the same zeros, x = 4 and x = 4
f(x) = x^2 - 8x + 16 would be the function with a double root at 4

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3 years ago

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TOY has coordinates T (-3, 4), O (-4, 1) and Y (-2, 3). a translation maps point T to T' (-1, 1). find the coordinates of O' and

NeX [460]

Answer:

Answer: The correct option is (A) O' (−2, −2); Y' (0, 0).

Step-by-step explanation: Given that the co-ordinates of the vertices of ΔTOY are T(−3, 4), O (−4, 1), and Y (−2, 3). A translation maps point T to T' (−1, 1).

We are to find the co-ordinates of the points O' and Y'.

The given transformation from T to T' is

T(−3, 4) ⇒ T' (−1, 1).

Let, (−3 + x, 4 + y) = (-1, 1).

So,

and

That is, the transformation rule is

(a, b) ⇒ (a+2, b-3).

Therefore,

co-ordinates of O' are (-4+2, 1-3) = (-2, -2),

and

co-ordinates of Y' are (-2+2, 3-3) = (0, 0).

Thus, the required co-ordinates of O' and Y' are (-2, -2) and (0, 0) respectively.

Option (A) is correct.

8 0

3 years ago

Which of the following expressions is a factor of the polynomial x 2 +3/2x-1

Elodia [21]

Answer:

$\large \boxed{\sf \ \ \ (x+2)(x-\dfrac{1}{2}) \ \ \ }$

Step-by-step explanation:

Hello,

let's solve

$x^2+\dfrac{3}{2}x-1=0\\\\ 2x^2+3x-2=0 \ \text{multiply by 2}\\\\\\$

$\Delta=b^2-4ac=9+4*2*2=9+16=25$

There are two solutions

$x_1=\dfrac{-3-\sqrt{25}}{4}\\\\x_1=\dfrac{-3-5}{4}\\\\x_1=\dfrac{-8}{4}\\\\\boxed{x_1=-2}$

And

$x_2=\dfrac{-3+\sqrt{25}}{4}\\\\x_2=\dfrac{-3+5}{4}\\\\x_2=\dfrac{2}{4}\\\\\boxed{x_2=\dfrac{1}{2}}$

So we can write

$x^2+\dfrac{3}{2}x-1\\\\=(x-x_1)(x-x_2)\\\\=(x+2)(x-\dfrac{1}{2})$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0

3 years ago

Please help don’t understand

const2013 [10]

The first one is congruent by ASA. The second does not have to be congruent, we don’t know enough.

5 0

3 years ago