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3 years ago

Simplify.

\sqrt{ \frac{4x^{2} }{3y} }" alt=" \sqrt{ \frac{4x^{2} }{3y} }" align="absmiddle" class="latex-formula">
Show your work.

Mathematics

1 answer:

Burka [1]3 years ago

4 0

$\sqrt{ \frac{4 x^{2} }{3y} } = \sqrt{ \frac{2^{2} x^{2}}{3y} } = \frac{ \sqrt{2^{2} x^{2}} }{ \sqrt{3y} } = \frac{2x}{ \sqrt{3y}}$

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M(4, 2) is the midpoint of . The coordinates of S are (6, 1). What are the coordinates of R? (8, 0) (2, 3) (1.5, 3) (5, 1.5)

faltersainse [42]

Assuming M is midpoint of RS

the midpoint formula is

x term of the midpoint is average of the x value of the 2 points
y term of the midpoint is average of the y value of the 2 points

if R is (x,y)

S is (6,1)

x value of midpoint is 4
x value of S is 6
x value os R is x

so
average of x and 6 is 4,
(x+6)/2=4,
x+6=8,
x=2

y value of midoint is 2
y value of S is 1
y value of R is y

average of 1 and y is 2
(y+1)/2=2
y+1=4
y=3

the midpoint is (2,3)

6 0

3 years ago

What is 5/8 - 4/7 but simplified? i dont know how to do it

SashulF [63]

Answer:3/56

Step-by-step explanation:

5/8-4/7

(35-32)/56

3/56

3 0

3 years ago

Find the sum of the positive integers less than 200 which are not multiples of 4 and 7

taurus [48]

Answer:

$12942$ is the sum of positive integers between $1$ (inclusive) and $199$ (inclusive) that are not multiples of $4$ and not multiples $7$ .

Step-by-step explanation:

For an arithmetic series with:

$a_1$ as the first term,
$a_n$ as the last term, and
$d$ as the common difference,

there would be $\displaystyle \left(\frac{a_n - a_1}{d} + 1\right)$ terms, where as the sum would be $\displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n)$ .

Positive integers between $1$ (inclusive) and $199$ (inclusive) include:

$1,\, 2,\, \dots,\, 199$ .

The common difference of this arithmetic series is $1$ . There would be $(199 - 1) + 1 = 199$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}$ .

Similarly, positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $4$ include:

$4,\, 8,\, \dots,\, 196$ .

The common difference of this arithmetic series is $4$ . There would be $(196 - 4) / 4 + 1 = 49$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $7$ include:

$7,\, 14,\, \dots,\, 196$ .

The common difference of this arithmetic series is $7$ . There would be $(196 - 7) / 7 + 1 = 28$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}$

Positive integers between $1$ (inclusive) and $199$ (inclusive) that are multiples of $28$ (integers that are both multiples of $4$ and multiples of $7$ ) include:

$28,\, 56,\, \dots,\, 196$ .

The common difference of this arithmetic series is $28$ . There would be $(196 - 28) / 28 + 1 = 7$ terms. The sum of these integers would thus be:

$\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}$ .

The requested sum will be equal to:

the sum of all integers from $1$ to $199$ ,
minus the sum of all integer multiples of $4$ between $1\!$ and $199\!$ , and the sum integer multiples of $7$ between $1$ and $199$ ,
plus the sum of all integer multiples of $28$ between $1$ and $199$ - these numbers were subtracted twice in the previous step and should be added back to the sum once.