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3 years ago

Answer in slope intercept form (y = Mx + b)

Mathematics

1 answer:

denpristay [2]3 years ago

5 0

The answer is y= 1x + 12

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What is bigger 3 yards or 10 feet

Studentka2010 [4]

10 feet is bigger than 3 yards I think not so positive

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3 years ago

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Write a equation that is parallel to 5x-y=4

ozzi

Let's have this equation equal y first. All we have to do is add y to both sides and subtract 4 from both sides.
5x-4=y
Now, to get a parallel line, the y intercept has to change. The y intercept (the -4 in the equation) can be an infinite amount of number but -4.
Let's choose 6. We would have 5x+6=y or you could put it in the way the other equation was, 5x-y=-6.

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3 years ago

HELPP ASAP PLS what is m A) 120° B) 75° C) 80° D) 85°

MariettaO [177]

The straight angle ;
60 + x = 180
So x = 120

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2 years ago

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Write this number in a standard notation 13,005

maks197457 [2]

1.3005 x 10^4
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3 years ago

Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that s

WINSTONCH [101]

Answer: $\bold{c=\dfrac{1+\sqrt{19}}{3}\approx1.8}$

Step-by-step explanation:

There are 3 conditions that must be satisfied:

f(x) is continuous on the given interval
f(x) is differentiable
f(a) = f(b)

If ALL of those conditions are satisfied, then there exists a value "c" such that c lies between a and b and f'(c) = 0.

f(x) = x³ - x² - 6x + 2 [0, 3]

1. There are no restrictions on x so the function is continuous $\checkmark$

2. f'(x) = 3x² - 2x - 6 so the function is differentiable $\checkmark$

3. f(0) = 0³ - 0² - 6(0) + 2 = 2

f(3) = 3³ - 3² - 6(3) + 2 = 2

f(0) = f(3) $\checkmark$

f'(x) = 3x² - 2x - 6 = 0

This is not factorable so you need to use the quadratic formula:

$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(-6)}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{4+72}}{2(3)}\\\\\\.\quad =\dfrac{2\pm \sqrt{76}}{2(3)}\\\\\\.\quad =\dfrac{2\pm 2\sqrt{19}}{2(3)}\\\\\\.\quad =\dfrac{1\pm \sqrt{19}}{3}\\\\\\.\quad \approx\dfrac{1+4.4}{3}\quad and\quad \dfrac{1-4.4}{3}\\\\\\.\quad \approx1.8\qquad and\quad -1.1$

Only one of these values (1.8) is between 0 and 3.

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3 years ago