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3 years ago

PLEASE HELP.⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Mathematics

1 answer:

balu736 [363]3 years ago

3 0

No incorrect. Would only be the origin at 1/2 G if H was (6,7) and I was ( -6,-7)

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√8x^2 divided by √2x

Ghella [55]

Question: Simplify $\frac{\sqrt{8x^2}}{\sqrt{2x}}$ .

First off, you <em>never</em> want square roots on the bottom of a fraction, so let's multiply both sides by $\sqrt{2x}$ to get rid of it. All we have to do from there is simplify!

$\frac{\sqrt{8x^2}}{\sqrt{2x}}*\frac{\sqrt{2x}}{\sqrt{2x}}=\frac{\sqrt{8x^2}*\sqrt{2x}}{2x}}=\frac{\sqrt{16x^3}}{2x}=\frac{4x\sqrt{x}}{2x}=\boxed{2\sqrt{x}}$

7 0

3 years ago

Read 2 more answers

The two lines, P and Q, are graphed below: Line P is drawn by joining ordered pairs negative 8,15 and 6, negative 12. Line Q is

andreyandreev [35.5K]

Solution:

Keep in mind ,

Equation of line joining two points (a,b) and (p,q) is given by :

$\frac{q-b}{p-a}=\frac{y-b}{x-a}$

Equation of line P which is obtained by joining (- 8,15) and (6, - 12) is given by:

$\frac{y-15}{x+8}=\frac{-12-15}{6+8}\\\\ 14(y-15)=-27(x+8)\\\\ 14 y -210= -27 x - 216\\\\ 27 x+14 y+6=0$

Equation of line Q which is obtained by joining (4,16) and (-9, 10) is given by:

$\frac{y-16}{x-4}=\frac{16-10}{4+9}\\\\ 13(y-16)=6(x-4)\\\\ 13 y -208= 6 x - 24\\\\ 6 x-13 y+184=0$

Equation of line P and Q are

27 x+14 y+6=0-------(1)× 2

6 x-13 y+184=0-------(2)× 9

54 x + 2 8 y+12=0---(1)

54 x -117 y +1656=0----(2)

(1) - (2)

145 y= 1644

y=11.33,

27 x+14 y+6=0-------(1)×13

6 x-13 y+184=0-------(1)×14

351 x + 182 y + 78=0-----(1)

84 x - 182 y +2576=0----(2)

(1) + (2)

435 x + 2654=0

x = - 6.11

So, solution set is (-6.11, 11.33)

Option (C) is true. (−2, 4), because this point makes both the equations incorrect.

8 0

3 years ago

Solve 2cos ²y -siny -1=0 for 0° ≤y≤360°

natima [27]

$\displaystyle\bf\\2cos^2y -sin\,y -1=0~~~for~~0^o\leq y \leq360^o\\\\cos^2y=1-sin^2y\\\\2(1-sin^2y) -sin\,y -1=0\\\\2-2sin^2y-sin\,y-1=0\\\\-2sin^2y-sin\,y+2-1=0\\\\-2sin^2y-sin\,y+1=0~~~\Big|\times(-1)\\\\2sin^2y+sin\,y-1=0$

$\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}$

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3 years ago

A consumer advocacy group wants to determine whether there is a difference between the proportions of the two leading automobile

podryga [215]

Answer:

The answer is "0.3206".

Step-by-step explanation:

$H_0: p_1 = p_2\\\\H_a: p_1 \neq p_2\\\\\hat{p_1} = \frac{X_1}{N_1} = \frac{53}{400} = 0.1325\\\\\hat{p_1}= \frac{X_2}{N_2} = \frac{78}{500}= 0.156\\\\\hat{p} = \frac{(X_1 + X_2)}{(N_1 + N_2)} = \frac{(53+78)}{(400+500)} = 0.1456$