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slava [35]
3 years ago
10

In the right triangle shown, m B с 12 45 A How long is BC?

Mathematics
1 answer:
olga55 [171]3 years ago
5 0

Answer:

BC = \sqrt{12^{2} +12^{2} }=12\sqrt{2}

Step-by-step explanation:

if the triangle is ABC and have part 12

and right angle from B and have an angle 45 then the third angle is 45 too

then we have another part is 12

and the BC = \sqrt{12^{2} +12^{2} }=12\sqrt{2}

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What is the slope of the line (9,-6) and (-6- -9)?
Vlada [557]

Answer:

1/5

Step-by-step explanation:

Slope formula = (y2-y1)/(x2-x1)

((-6)- (-9))/ ((9)-(-6)

(-6 + 9)/ (9 + 6)

3/15

1/5

6 0
3 years ago
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What is 20.6 divided by 9.7? (Include remainder)
andre [41]
20.6 divided by 9.7 is 2.12371134
5 0
3 years ago
This??? What is wrong with it?
ELEN [110]

Answer:

15.8 sq. in. of paper will be required.

Step-by-step explanation:

The problem is that a drinking cup does not have a cover, so only the lateral surface area counts.

I.e. We need only the first term.

A = pi r l = pi * 1.5 * sqrt(3^2+1.5^2)

= 15.81 sq. in.

4 0
3 years ago
Helppp how do u do 12
Delicious77 [7]
B is the correct answer



\frac{x}{100}  =  \frac{53}{106}  \\ x =  \frac{53 \times 100}{106}  =  \frac{5300}{106}  = 50




good luck
4 0
3 years ago
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For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

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

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

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0
3 years ago
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