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s2008m [1.1K]
3 years ago
11

Give the slope of line 6x+ 1/2 y= -3

Mathematics
2 answers:
fredd [130]3 years ago
8 0

Answer:

-12

Step-by-step explanation:

leva [86]3 years ago
4 0
The answer is -> -12
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steve made a mistake in writing a two step equation. Instead of multiplying by 3 and adding 7, he multiplied by 7 and added 3 to
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Answer:

The correct answer of the problem is 1.

Step-by-step explanation:

Let the original number on which the wrong procedure is applied is x.

Now, Steve multiplied by 7 and then added 3 to get the answer of - 11.

So, 7x + 3 = - 11

⇒ 7x = - 14

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Now, the correct step to get the answer is multiplying by 3 and adding 7.

Hence, the correct answer of the problem is (- 2) × 3 + 7 = - 6 + 7 = 1 (Answer)

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4 years ago
Dina and Masha started out on a 10 km bike path at the same time. When Dina reached the end of the 10 km, Masha still had 2 km l
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What is the area of the Largest square (BRAINLIST WILL BE GIVE IF ANSWER CORRECT)
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Using pythagoras theorem, it states that 36+64= area of the unknown

3 0
3 years ago
Simplify the expression 13+|-21-(-4)²|
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13+|-21-(-4)^2| \\ \\ 13 + |-21-4^2| \\ \\ 13 + |-21-16| \\ \\ 13 + |-37| \\ \\ 13 + 37 \\ \\ 50 \\ \\ Answer: \fbox {50}
7 0
4 years ago
PLEASE HELP ME! I really need help y'all...
ikadub [295]

just a quick clarification, tis usually -4.9 and that's a rounded number to reflect earth's gravity on an object in motion, but -5 is close enough :)

\bf ~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf h(x)=-5(\stackrel{\mathbb{F~O~I~L}}{x^2-8x+16})+180\implies h(x)=-5x^2+40x-80+180 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x)=-5x^2+\stackrel{\stackrel{v_o}{\downarrow }}{40} x+\stackrel{\stackrel{h_o}{\downarrow }}{\boxed{100}}~\hfill

8 0
4 years ago
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