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adelina 88 [10]

2 years ago

8

Which of the sets of ordered pairs represents a function? (5 points) A = {(2, −2), (5, −5), (−2, 2), (−5, 5)} B = {(4, 2), (4, −

2), (9, 3), (9, −3)} Group of answer choices Only A Only B Both A and B Neither A nor B

1 answer:

ELEN [110]2 years ago

6 0

Answer:

A

Step-by-step explanation:

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Help!<br> JH is a mid-segment of KLM. Find the value of x.

USPshnik [31]

JH = mid-segment of triangle KLM.

ML = 22

(1/2)(ML) = 11

JH = (1/2)(ML) = x

So, x = 11.

7 0

3 years ago

If your initial principal is $243 and your principal in 2 years is $1,359.19, how much interest did you earn?

jekas [21]

Answer: $1,116.19

Step-by-step explanation:

If you invested $243 and this increased to $1,359.19 in two years, the difference between your investment and its current worth is the interest:

= Current principal - Initial principal

= 1,359.19 - 243

= $1,116.19

5 0

3 years ago

A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556

MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

The formula to find the standard deviation:-

$\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}$

The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.

Then, $\overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560$

Now, $\sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76$

Then, $\sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76$

Hence, the standard deviation of his measurements = 2.76 g

6 0

4 years ago

HELPPP what is 2x - 5 by the Square root of 9

ioda

2x-15 is the answer for 2x-5 sq 9

4 0

3 years ago

Rewrite the expression in the form x^n

Charra [1.4K]

Answer:

$\sqrt[3]{x}$

Step-by-step explanation:

$\sqrt[4]{ \frac{ {x}^{2} }{x ^{ \frac{2}{3} } } } = \frac{ \sqrt[4]{ {x}^{2} } }{ \sqrt[4]{ \sqrt[3]{ {x}^{2} } } } = \frac{x ^{ \frac{2}{4} } }{(x ^{ \frac{2}{3} } ) ^{ \frac{1}{4} } } \\ \\ = \frac{ {x}^{ \frac{1}{2} } }{ {x}^{ \frac{2}{12} } } = \frac{ {x}^{ \frac{1}{2} } }{ {x}^{ \frac{1}{6} } } = {x}^{ \frac{1}{2} - \frac{1}{6} } = {x}^{ \frac{1}{3} } = \sqrt[3]{x}$

I hope I helped you^_^

3 0

3 years ago