PLEASE HELP ASAP!

Mathematics

2 answers:

puteri [66]3 years ago

5 0

The answer is B

To figure this out, all you have to do is put 2/3 into a ratio with 6 so it looks like 2/3:6 now get rid of the fraction by multiplying both sides by 3/2 to make the left side 1 and the right side 4.

denis-greek [22]3 years ago

5 0

The answr is A because 6 divided by 2/3 is 9, and △ABC is smaller than △ DEF. Sorry if i'm wrong.

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If R is the total resistance for a parallel circuit with two resistors of resistances of r1 and r2 then 1/r=1/r1+1/r2 find the r

Allushta [10]

A little Algebra, and we're home free!

We are given R = 20 and r2 = 75. We need to find r1.

The formula is 1/R = 1/r1 + 1/r2. Plug-in:
1/20 = 1/r1 + 1/75
1/20 - 1/75 = 1/r1
0.0366 = 1/r1
r1 = 1/0.03666
r1 = 27.272

Our answer is r1 = 27 ohms.

6 0

3 years ago

Read 2 more answers

Find the following:

Butoxors [25]

Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ , if f(x) approaches L as x approaches 'a'.

(a)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\$

$\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

(b)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$