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3 years ago

Graph y=-2 and graph y=-x-4

Mathematics

1 answer:

Sati [7]3 years ago

4 0

On the y=-2, draw a horizontal line on the -2 of y. The second one, your points will be, (0,0) (-1,1)(-2,2)(-3,3) and draw a straight line through the whole graph through those points.

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Complete the solution of the equation. find the value of y when x equals 1. 9x +7y= -12

Taya2010 [7]

Answer:

$y=-3$

Step-by-step explanation:

Given Equation:

$9x +7y=-12\\$ y=? at x= 1

Putting the given value of 'x' in the equation:

$9(1) +7y=-12\\\\9+7y=-12\\\\7y=-12-9\\\\7y=-21\\\\y=-\frac{21}{7} \\y=-3$

At $x=1$ , the value of y is $y=-3$

6 0

3 years ago

Which best describes the composition of two functions

masha68 [24]

The composition of 2 functions is B

3 0

3 years ago

Divide write the answer in simplest form

Rom4ik [11]

Answer:

3 is the answer

1/3*9/1=9/3=3

Step-by-step explanation:

6 0

2 years ago

2(a³+b²+11)+1(3+a³+b+11)

Ann [662]

Answer:

$2(a^3+b^2+11)+1(3+a^3+b+11)=3a^3+2b^2+b+36$

Step-by-step explanation:

I assume that you need simplification of the given expression.

The given expression is:

$2(a^3+b^2+11)+1(3+a^3+b+11)$

Using distributive property and multiplying 2 inside the parenthesis and 1 inside the other parenthesis. This gives,

$2(a^3+b^2+11)=2\times a^3+2\times b^2+2\times 11\\2(a^3+b^2+11)=2a^3+2b^2+22\\\\1(3+a^3+b+11)=1\times 3+1\times a^3+1\times b+1\times 11\\1(3+a^3+b+11)=3+a^3+b+11=a^3+b+14$

Therefore, $2(a^3+b^2+11)+1(3+a^3+b+11)$ is equal to:

$2a^3+2b^2+22+a^3+b+14$

Now, combining like terms using the commutative property of addition, we get:

$=(2a^3+a^3)+2b^2+b+(22+14)\\=3a^3+2b^2+b+36$

Therefore, the simplified form is $3a^3+2b^2+b+36$

8 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 x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

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

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -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 f(x) is differentiable everywhere on its domain, except at the point x = -1.

4 0

3 years ago