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kondaur [170]
3 years ago
7

What is the slope?

Mathematics
1 answer:
Verizon [17]3 years ago
7 0
The slope is -2 or -2/1 
the x intercept is -8 
the y intercept is -4
the equation of the line in slope intercept form is y= -2x-4
the equation of the line in standard form is -2x-4y=0
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Convert a half to decimal
Ksivusya [100]

The answer would be  is 0.5

7 0
3 years ago
Read 2 more answers
Calculate the following limit:
aleksklad [387]
\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\
=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1


8 0
3 years ago
Please help with this problem
lawyer [7]

Answer:

624

Step-by-step explanation:

Plug into the formula, SA=bh+(s1+s2+s3)H.

H=14

s1=13

s2=13

s3=10

Hope that helps!

7 0
3 years ago
15/15 as a whole number
vladimir2022 [97]
Bruh it's 15. i mean if it is 15/15 then there is 15 things present yunno? idk but i hope this helped, have an amazing day :)
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20-%2011%20%3D%207%20-%20x" id="TexFormula1" title=" - 11 = 7 - x" alt=" - 11 = 7 - x" align=
Oksana_A [137]
X = 18 
if you need the steps i could give them to you just ask! 
Hope that helped!
:)
5 0
3 years ago
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