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Mashutka [201]
3 years ago
13

Y varies directly as the square of x when x=2 then y =16 find y when x=8

Mathematics
1 answer:
Leviafan [203]3 years ago
8 0

Answer:   y= 64  

Step-by-step explanation:

If y is 16 when  x is 2  This means if  x is 8 y will be 64.

Set up the proportional equation

\frac{2}{16}  = \frac{8}{y}     Cross multiply  

2y = 128

y= 64

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Annette [7]

Answer:

The two angles mentioned are alternate interior angles, so they are congruent.

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Step-by-step explanation:

If you draw off the flag you can see they are alternate interior and are congruent.

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6 0
3 years ago
What’s the answer plz hurry!
Svetllana [295]

Answer:

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7 0
3 years ago
Add the two expressions.
Ierofanga [76]
-3.2b + 9
by adding like terms together.
3 0
3 years ago
Read 2 more answers
Few math questions TIME IS RUNNING OUT!! PLEASE HELP!!!!
olya-2409 [2.1K]
1.

|5+6x|\ \textless \ 13\\\\-13\ \textless \ 5+6x\ \textless \ 13

so

\begin{cases}5+6x\ \textless \ 13\\5+6x\ \textgreater \ -13\end{cases}

Answers B and D.

2.

|x+9|\geq11\\\\x+9\ \textgreater \ 11\qquad\vee\qquad x+9\leq-11\\\\
\boxed{x\geq2\qquad\vee\qquad x\leq-20}

Answer D.

3.

The same as point 2. Answer D.

4.

Answer C.

3 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
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