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fomenos
3 years ago
13

How do you add/subtract radical operations? I know how to simply and stuff, but how do you do things like 4 square root of 2 plu

s 7 square root of 4?

Mathematics
1 answer:
LuckyWell [14K]3 years ago
8 0
U get a calculator and u put 4^2+7^4 n u put it in the calculator but the pic shows what a sqaure root looks like n u see the button with the x2 u can press that for when u put the 4 square root 2 then plus 7 n then u hit the button that has this symbol ^ n hit 4 then equal an itll give u the answer which is 2417.

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Triangle ABC is translated according to the rule (x, y) → (x + 2, y – 8). If the coordinates of the pre-image of point B are
yuradex [85]
Ans: (6,-13) 

Rationale: 

Simply take your pre-image (point before applying transformation) of point B (4,-5) and apply the transformation to each point. Therefore, (x,y) = (4+2,-5-8) = (6,-13)
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These need to be answered in 15 minutes. Here are some math questions...
OLEGan [10]

1) 2, greater than (>)

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2) 1, less than (<)

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4 years ago
Can anyone help me with calculus??
gogolik [260]

1. If f(x)=(x+1)^4, then f'(x)=4(x+1)^3. So f'(1)=32.

2. With x^2+y^2=1, we differentiate once with respect to x and get

\dfrac{\mathrm d}{\mathrm dx}[x^2+y^2]=\dfrac{\mathrm d}{\mathrm dx}1

2x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac xy

Differentiate again with respect to x and we get

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{y-x\frac{\mathrm dy}{\mathrm dx}}{y^2}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{y+\frac{x^2}y}{y^2}=-\dfrac{y^2+x^2}{y^3}=-\dfrac1{y^3}

(where y\neq0).

3. Check the one-side limits where the pieces are split. For f to be continuous everywhere, we need

\displaystyle\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)

In the first case, we have

\displaystyle\lim_{x\to-1^-}f(x)=\lim_{x\to-1}x+2=1

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and f(-1)=1, so it's continuous here.

In the second case, we have

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^2=1

\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}3-x=2

so f is discontinuous at x=1.

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nikklg [1K]

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Step-by-step explanation:

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