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Nataliya [291]
3 years ago
12

Pls answer any that you can!!!

Mathematics
1 answer:
klio [65]3 years ago
6 0

Answer:

12, 45, 1.8, -9, 18

Step-by-step explanation:

up up down down left right left right  b a start

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Determine the slope-intercept form of the equation of the line parallel to y = x + 11 that passes through the point (–6, 2). y =
Vika [28.1K]

Answer:

y = x+12

Step-by-step explanation:

Parallel => <u><em>This means it has the same slope as this one.</em></u>

Slope = m = 1

Now,

Point = (x,y) = (-6,2)

So, x = -6, y = 2

<u><em>Putting this in slope intercept form to get b</em></u>

y = mx+b

=> 2 = (1)(-6) + b

=> b = 2+6

=> b = 8

<u><em>Now putting m and b in the slope-intercept form to get the required equation:</em></u>

=> y = mx+b

=> y = x+12

6 0
3 years ago
What is the value of x?
Helga [31]

2x+20=3x-30

x=50

2x+20=3x-30

x=50

2x+20=3x-30

x=50

2x+20=3x-30

x=50

7 0
3 years ago
Solve |p + 2| = 10Solve |p + 2| = 10<br><br> {-12}<br> {-8, 8}<br> {-12, 8}
valentinak56 [21]

By definition, we have

|p+2| = \begin{cases} p+2 &\text{ if } p+2 \geq 0 \\-p-2 &\text{ if } p+2 < 0 \end{cases}

So, we have to solve two different equations, depending of the possible range for the variable. We have to remember about these ranges when we decide to accept or discard the solutions:

Suppose that p+2\geq 0 \iff p \geq -2

In this case, the absolute value doesn't do anything: the equation is

p+2 = 10 \iff p = 10-2 = 8

We are supposing p \geq -2, so we can accept this solution.

Now, suppose that p+2 < 0 \iff p < -2. Now the sign of the expression is flipped by the absolute value, and the equation becomes

-p-2 = 10 \iff -p = 12 \iff p = -12

Again, the solution is coherent with the assumption, so we can accept this value as well.

3 0
3 years ago
This one too please!!!!!!!!!
Vinvika [58]
X= 0 and 1 are the answers
7 0
3 years ago
Simplify the rational expression.  State any restrictions on the variable.
AveGali [126]

Answer:

The last choice.

Step-by-step explanation:

\dfrac{k^2-k-2}{k^2-4k-5}=\dfrac{(k+1)(k-2)}{(k+1)(k-5)}=\dfrac{k-2}{k-5}

The value of k must be restricted from being one that makes the original denominator zero. Hence k ≠ -1 or 5.

7 0
3 years ago
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