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Alecsey [184]
3 years ago
8

Compute: 11^2−(12−4)^2+3=

Mathematics
2 answers:
masha68 [24]3 years ago
8 0

Answer:

60

Step-by-step explanation:

klasskru [66]3 years ago
7 0
Answer:
60


Do 11^2 (121) then subtract that by 8^2 (64) and add 3
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Isolated triangle 68 perimeter equal sides more than 10 ft use x for the shorter side formula
Ugo [173]

Answer:

Step-by-step explanation:

Find the perimeter of an isosceles triangle whose equal sides have a size of 10 m each and the angle between them equal to 30°. We need to know all sides in order to find the perimeter of this triangle. Let x be the base of this isosceles triangle.

6 0
2 years ago
Simplify the expression.<br><br><br> 5.3x - 9 + 7.6x
babunello [35]

Answer:

12.9x-9

Step-by-step explanation:

First you have to combine the like terms which are 5.3x + 7.6x= 12.9x. Then you just rewrite the equation 12.9x-9. You just leave the 9 by it self beacuse thers no other like terms.

7 0
2 years ago
Read 2 more answers
Solve the equation sine Ф=0.6792 for 0°≤Ф≤360
Sergio [31]

Answer:

42.78⁹, 137.22⁹.

Step-by-step explanation:

sine Ф=0.6792

Angle Ф in the first quadrant = 42.78 degrees.

The sine is also positive in the second quadrant so the second solutio is

180 - 42.78

= 137.33 degres.

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20-7x%2B8%5Cleq%205" id="TexFormula1" title="x^{2} -7x+8\leq 5" alt="x^{2} -7x+8\l
nikdorinn [45]

Answer:

x^2-7x+3=0

Step-by-step explanation:

5 0
3 years ago
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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