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iogann1982 [59]
3 years ago
8

The intersection of two planes is a POINT PLANE LINE LINE SEGMENT

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
3 0
The intersection of two planes is a line.
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The results of a test of a metal are shown in the table, with strain in meters per meter, andstress in units called megapascals,
shusha [124]

a.

b.

The metal obey this law for values of strain until 0.05, where we have a linear relationship (each increase of 0.01 in the strain causes an increase of 100 in the stress). After this point, we don't have a linear relationship anymore.

c. Since an increase of 0.01 in the strain causes an increase of 100 in the stress, the slope is:

m=\frac{100}{0.01}=10000

Now, calculating the coefficient b (y-intercept), we have:

\begin{gathered} (0.01,100)\colon \\ 100=0.01\cdot10000+b \\ 100=100+b \\ b=0 \end{gathered}

So the equation is:

y=10000x

d.

The maximum value of stress is 560, and occurs at strain = 0.07.

8 0
8 months ago
What are the intercepts for the equation? Select all that apply. x + y = 4 *
olchik [2.2K]

Answer:

  y intercept (0, 4)

  x intercept (4, 0)

I hope this is good enough:

3 0
3 years ago
Which table represents the equation y = -2x + 17?
Alexeev081 [22]
<h3>Answer: Choice A</h3>

Explanation:

Each table has x = 10 in it. Plug this value into the given equation.

y = -2x+17

y = -2*10+17

y = -20+17

y = -3

The input x = 10 leads to the output y = -3. Table A shows this in the middle-most row. So that's why choice A is the answer. The other tables have x = 10 lead to y values that aren't -3 (eg: choice D has x = 10 lead to y = 12), so we can rule them out.

7 0
2 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
2 years ago
What is the slope-intercept equation for the line below?
Kryger [21]
It would be at 35 when you work all then problems together
8 0
3 years ago
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