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saw5 [17]
3 years ago
8

I need help really bad pls help

Mathematics
2 answers:
Ahat [919]3 years ago
8 0
Answer: x<=1

Explanation: you first move the constant to the right, making the equation turn into -13x>=-1-12. Next, you calculate -1-12 which is -13, and plug that into the equation to make -13x>=-13. Last but not least, you divide -13 on both sides to leave x by itself, and flip the inequality sign. Then you get your answer: x<=1. Hope this helps! :)
Damm [24]3 years ago
5 0
Answer: C
Have a good day :D
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What is the solution for <br> 10+2w&gt;=22 or 5w-8&gt;-12
Marina CMI [18]

Answer:

no solution

Step-by-step explanation:

10+2w>=22 or 5w-8>-12

10-2w>=22 or (5w-2w)-8>-12

10>=22 or 3w+8>12

18>=22 or 3w>12

18/3>=22 or 3w/3>12

6>=22 or w>12

3 0
2 years ago
Which inequality is equivalent to 7x-2y&gt;8?
Crazy boy [7]
7x-2y\ \textgreater \ 8  \\  7x-8 \ \textgreater \ 2y   /:2  \\   \frac{7}{2} x -4\ \textgreater \ y  \\  y\ \textless \  \frac{7}{2} x-4
4 0
4 years ago
Because of their connection with secant​ lines, tangents, and instantaneous​ rates, limits of the form ModifyingBelow lim With h
Gre4nikov [31]

Answer:

\dfrac{1}{2\sqrt{x}}

Step-by-step explanation:

f(x) = \sqrt{x} = x^{\frac{1}{2}}

f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}

We use binomial expansion for (x+h)^{\frac{1}{2}}

This can be rewritten as

[x(1+\dfrac{h}{x})]^{\frac{1}{2}}

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}

From the expansion

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots

Setting x=\dfrac{h}{x} and n=\frac{1}{2},

(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots

=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots

Multiplying by x^{\frac{1}{2}},

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots

The limit of this as h\to 0 is

\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}} (since all the other terms involve h and vanish to 0.)

8 0
3 years ago
Please help me I need ASAP please please help me please please help please
Kruka [31]

Answer:

3

Step-by-step explanation:

i am so sorry if i am not correct

5 0
3 years ago
Read 2 more answers
What is 2.3w-7+8.1-3w
stepan [7]
2.0 w-1.1
2.3 w -3w = 2.0 w , -7+8.1 =1.1
8 0
4 years ago
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