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3 years ago

Hey guys it's been a while can you just help me with these two problems?

Mathematics

2 answers:

lutik1710 [3]3 years ago

7 0

Answer:

Your already right on both of them lol

Step-by-step explanation:

Pls brainliest thx! :D

siniylev [52]3 years ago

5 0

Answer:

you're right

Step-by-step explanation:

it's b, and a!

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Another derivatives question, can I get help please?

alex41 [277]

$\bf \textit{de}\textit{finition of a derivative as a limit}\\\\
\lim\limits_{h\to 0}~\cfrac{f(x+h)-f(x)}{h}\\\\
-------------------------------\\\\
f(x)=5x^2-6x-3
\\\\\\
\lim\limits_{h\to 0}~\cfrac{[5(x+h)^2-6(x+h)-3]~~-~~(5x^2-6x-3)}{h}
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\lim\limits_{h\to 0}~\cfrac{[5(x^2+2xh+h^2)-6(x+h)-3]~~-~~5x^2+6x+3}{h}$

$\bf \lim\limits_{h\to 0}~\cfrac{\underline{5x^2}+10xh+5h^2\underline{-6x}-6h\underline{-3}~~~\underline{-5x^2+6x+3}}{h}
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\lim\limits_{h\to 0}~\cfrac{10xh+5h^2-6h}{h}\implies \lim\limits_{h\to 0}~\cfrac{\underline{h}(10x+5h-6)}{\underline{h}}
\\\\\\
\lim\limits_{h\to 0}~10x+5(0)-6\implies \lim\limits_{h\to 0}~10x-6$

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3 years ago

The function f is represented by f(x)=3x2^x. The function is represented by the table below.

Oksanka [162]

I suppose it’s -16fxg

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3 years ago

Solve each equation by completing the square. If necessary, round to the nearest hundredth.

Ksenya-84 [330]

$1.)\\ \\ b^2+10b=75\\ \\ b^2+10b-75=0\\ \\\Delta = b^{2}-4ac = 10^{2}-4*1*(75)=100+300=400 \\ \\\sqrt{\Delta }=\sqrt{400}=20$

$b_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-10-20}{2}=\frac{-30}{2}=-15\\ \\b_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{-10+20}{2}=\frac{10}{2}=5$

$2.)\\ \\ n^2-20n=-75\\ \\ n^2-20n+75=0\\ \\\Delta = b^{2}-4ac = (-20)^{2}-4*1*75=400-300=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$n_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{20-10}{2}=\frac{10}{2}=5\\ \\n_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{20+10}{2}=\frac{30}{2}=15$

$3.)\\ \\t^2+8t-9=0\\ \\\Delta = b^{2}-4ac = 8^{2}-4*1* (-9)=64+36=100 \\ \\\sqrt{\Delta }=\sqrt{100}=10$

$t_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-8-10}{2}=\frac{-18}{2}=-9\\ \\t_{2}=\frac{-b+\sqrt{\Delta }}{2a} \frac{-8+10}{2}=\frac{2}{2}=1$

$4.)\\ \\m^2-2m-8=0\\ \\\Delta = b^{2}-4ac = (-2)^{2}-4*1* (-8)=4+32=36\\ \\\sqrt{\Delta }=\sqrt{36}=6$

$m_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{2-6}{2}=\frac{-4}{2}=-2\\ \\m_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{2+6}{2}=\frac{8}{2}=4$

$5.)\\ \\ v^2+4v-2=0\\ \\\Delta = b^{2}-4ac = 4^{2}-4*1* (-2)=16+8=24\\ \\\sqrt{\Delta }=\sqrt{24}= \sqrt{4*6}=2\sqrt{6}$

$v_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{-4-2\sqrt{6}}{2}=\frac{2(-2-\sqrt{6}}{2}= -2-\sqrt{6}\approx -2-2,45\approx -4,45\\ \\v_{2}=\frac{-b+\sqrt{\Delta }}{2a}= \frac{-4+2\sqrt{6}}{2}=\frac{2(\sqrt{6}-2}{2}= \sqrt{6}-2 \approx 2,45-2\approx 0,45$

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3 years ago

How to find if a function is positive or negative?

stiks02 [169]

Positive because a function is beetween and past

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3 years ago

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R = sqrt(3,6^2+(11/2)^2) = 6,6

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4 years ago

Hey guys it's been a while can you just help me with these two problems?​

Hey guys it's been a while can you just help me with these two problems?