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sveta [45]
3 years ago
10

If angle X has a measure of 40 degrees and angle Y is vertical to angle X what is the measure of angle Y?

Mathematics
2 answers:
ad-work [718]3 years ago
7 0
Angle y measure would most likely be 20 degrees
erica [24]3 years ago
6 0
If an angle is 70 degrees and there is an angle vertical to it, then that other angle also equals 70 degrees. Therefore, if angle X equals 40 degrees, then angle Y also equals 40 degrees. Hope this helps.
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The area of a rectangle is found by multiplying its length and width together. Calculate the area, in square feet, of the room y
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Answer:

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Step-by-step explanation:

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Answer:

Distance to from house to school to community center

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Step-by-step explanation:

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3 years ago
The quotient of (x^3+3x^2-4x-12)/(x^2+5x+6)
Elis [28]

Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}

Now, we have to find the quotient of the above expression.

So, here we go:

Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)

Now, we will divide the above simplest form with g(x):

\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2

Hence, the quotient is (x-2).


6 0
3 years ago
Read 2 more answers
Henry is using cardboard to build a box in the shape of a rectangular prism. A net of the box is shown.
Brut [27]

Answer:240 square feet

Step-by-step explanation:

5 0
3 years ago
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Given points A(-1, -2) and B(2, 4) where AP: BP=1:2, find the locus of point P.​
koban [17]

Answer:

x^2 + 4x + y^2 +8y  =  0

Step-by-step explanation:

Given

A = (-1,-2)

B = (2,4)

AP:BP = 1 : 2

Required

The locus of P

AP:BP = 1 : 2

Express as fraction

\frac{AP}{BP} = \frac{1}{2}

Cross multiply

2AP = BP

Calculate AP and BP using the following distance formula:

d = \sqrt{(x - x_1)^2 + (y - y_1)^2}

So, we have:

2 * \sqrt{(x - -1)^2 + (y - -2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}

2 * \sqrt{(x +1)^2 + (y +2)^2} = \sqrt{(x - 2)^2 + (y - 4)^2}

Take square of both sides

4 * [(x +1)^2 + (y +2)^2] = (x - 2)^2 + (y - 4)^2

Evaluate all squares

4 * [x^2 + 2x + 1 + y^2 +4y + 4] = x^2 - 4x + 4 + y^2 - 8y + 16

Collect and evaluate like terms

4 * [x^2 + 2x + y^2 +4y + 5] = x^2 - 4x + y^2 - 8y + 20

Open brackets

4x^2 + 8x + 4y^2 +16y + 20 = x^2 - 4x + y^2 - 8y + 20

Collect like terms

4x^2 - x^2 + 8x + 4x + 4y^2 -y^2 +16y + 8y  + 20 - 20 =  0

3x^2 + 12x + 3y^2 +24y  =  0

Divide through by 3

x^2 + 4x + y^2 +8y  =  0

3 0
3 years ago
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