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PSYCHO15rus [73]
3 years ago
12

Which is not an equation of the line going through (6, 7) and (2,-1)?

Mathematics
1 answer:
atroni [7]3 years ago
3 0

Answer:

B. y-1=2(x+2)

Step-by-step explanation:

be the points  (6,7) and (2,-1)

replacing 7-1=2(6+2) → 6≠16 and -1-1=2(2+2) → -2≠8

for all equations the equalities are maintained

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Answer:

3 hours and 45 minutes assuming its from am to pm

Step-by-step explanation:

Hope This Helps

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2 years ago
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Jake tossed a paper cup 50 times and recorded how it landed. The table shows the results:
Dmitry [639]
Hello!

Here are 6 steps to Help You Out!

1. 3 divided by 50 = .06

2. Then if you like you can turn it in to a percent not sure how the answer is supposed to look for example:

<span><span>3. 3/50</span>=<span>6/100</span>=6%

</span>4. 7 divided by 50 - .14

5. so 0.06 for the first one

6. And 40 divided by 50 = .8

====>So as we can see, o<span>pen side up .06 closed side up .14 and  on its side .8 or open 6% closed 14% and side 80% chance.<====

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8 0
3 years ago
Solve: 14 = k(0.4) k = _____
KatRina [158]
<h2><u>k = 35</u></h2>

Let's solve your equation step-by-step.

14=k(0.4)

Step 1: Simplify both sides of the equation.

14=0.4k

Step 2: Flip the equation.

0.4k=14

Step 3: Divide both sides by 0.4.

\frac{0.4k}{0.4} = \frac{14}{0.4}

<h3><u>k=35</u></h3>

4 0
3 years ago
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What single transformación maps ABC onto ABC ?
Thepotemich [5.8K]

Answer:

a reflection across y-axis then translation of 1 unit right and 2 units up.

a clockwise rotation of 180° about the origin then a translation of 1 unit right and 3 units up

a reflection across y-axis then translation of 1 unit right and 1 unit down

a reflection across y = x then a positive rotation of 270° about the origin

Step-by-step explanation:

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3 years ago
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Use the four-step definition of the derivative to find f'(x) if f(x) = −4x^3 −1.
guapka [62]

\stackrel{de finition \textit{ of a derivative as a limit}}{\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{[-4(x+h)^3-1]~~ - ~~[-4x^3-1]}{h} \\\\\\ \cfrac{[-4(x^3+3x^2h+3xh^2+h^3)-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{[-4x^3-12x^2h-12xh^2-4h^3-1]~~ ~~+4x^3+1}{h} \\\\\\ \cfrac{-4x^3-12x^2h-12xh^2-4h^3-1+4x^3+1}{h}\implies \cfrac{-12x^2h-12xh^2-4h^3}{h}

\cfrac{h(-12x^2-12xh-4h^2)}{h}\implies -12x^2-12xh-4h^2 \\\\\\ \lim\limits_{h\to 0}~-12x^2-12xh-4h^2\implies \lim\limits_{h\to 0}~-12x^2-12x(0)-4(0)^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \lim\limits_{h\to 0}~-12x^2~\hfill

7 0
2 years ago
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