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3 years ago

Use the function to fill in the blanks in the table.

Mathematics

1 answer:

S_A_V [24]3 years ago

5 0

Answer:

gnrg

Step-by-step explanation:

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1. 800 divided by 4

grigory [225]

I did the work for you. I also showed my work. Hope it helped

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3 years ago

Please help please help

koban [17]

35. is c : square root of 64 is 8 and the square root of 8 can be simplifed to 4 x 2, square root of 4 is 2

36 is d : Denominator of the fraction represents what it is being rooted by and the numerator is what the regular number is squared by.

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3 years ago

Read 2 more answers

0.35 in expanded form

Diano4ka-milaya [45]

Answer:0.3+0.05

Step-by-step explanation:

7 0

3 years ago

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Chuge%5Cmathfrak%5Cpurple%7Bplease%20%5C%3A%20solve..%7D%20%5C%5C%20%20%5C%5C%20%5Chuge%5Cma

Verdich [7]

Answer:

2²4

4²=16

4+16=20

hope it helps :)

8 0

3 years ago