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Andreyy89
3 years ago
11

I NEED HELPPP ASAP pleaseeeeeee

Mathematics
2 answers:
Komok [63]3 years ago
5 0

Answer:

Subtract 23 from both sides

Step-by-step explanation:

This is how algebra works, to get a variable by itself you have to do the opposite to both sides. Hope this helped.

maw [93]3 years ago
4 0

Answer:

It's D: subtract 23 from both sides of the equation because inverse operations  

Step-by-step explanation:

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Which polynomial represents the area?
murzikaleks [220]

Answer:

A is the Answer.

Step-by-step explanation:

You multiply x with 2x ( Gives you 2x^2 ) then -1 with 2x ( Gives you -2x ) then -8 with x ( Gives you -8x ) then -1 with-8 ( Gives you +8 ), finally you add the two x's ( not x^2 ). Your fibal answer will be 2x^2 - 10x +8.

Hope this helps.

5 0
4 years ago
Let a=<-4,-3,5> Find a unit vector in the same direction as having positive first coordinate.
Rashid [163]

Answer:

The value is u =  < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>

Step-by-step explanation:

From the question we are told that

The vector is a=<-4,-3,5>

Generally the unit vector is u = ay

Here y represent the y-coordinate

So

u =y

=> u =

Generally the resultant of a unit vector is 1

So

|u| = \sqrt{ (-4y)^2 + (-3y)^2 + (5y)^2} = 1

Hence

|u| = \sqrt{ 16y^2 + 9y^2 + 25y^2} =  1

Taking the square of both sides

16y^2 + 9y^2 + 25y^2 =  1

=> 50y^2 =  1

=> y =  \pm \frac{1}{\sqrt{50}}

=> y =  \pm \frac{1}{5 \sqrt{2}}

Rationalizing

=> y =  \pm \frac{\sqrt{2}}{10}

Given that the first coordinate is positive

y = - \frac{\sqrt{2}}{10}

Hence the unit vector is

u =

=> u =  < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>

4 0
3 years ago
The scale used on a map is 5 millimeters represents 32 kilometers. How many millimeters represent 160 kilometers?
Anika [276]

Use the proportion formula:

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Dvinal [7]

Answer:

The vertex of angle is S.

Hope this helps!!

7 0
4 years ago
-3(m-2n)=12;c someone please help
Sav [38]

-3(m-2n)=12\ \ \ \ \ |:(-3)\\\\m-2n=-4\\\\\text{Solve for m:}\\\\m-2n=-4\ \ \ \ |+2n\\\\\boxed{m=2n-4,\ n\in\mathbb{R}}\\\\\text{Solve for n:}\\\\m-2n=-4\ \ \ \ |-m\\\\-2n=-m-4\ \ \ \ |:(-2)\\\\\boxed{n=\dfrac{m+4}{2},\ m\in\mathbb{R}}

4 0
3 years ago
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