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puteri [66]
3 years ago
12

How do you solve number 27 and 28

Mathematics
1 answer:
Elanso [62]3 years ago
8 0

27) if a = 1/a

multiply both sides by a

a * a = 1/a * a

a^ 2 = 1

Answer is D) 1

28)

6 x 75 = 450

7 x 80  = 560

8 x 85 = 680

9 x 90 = 810

Total all students in class = 6 + 7 + 8 + 9 = 30 students

Total score of whole class:

450 + 560 + 680 + 810 = 2500

Average = 2500 /30 = 83.3 = 83 1/3

Answer is C) 83 1/3

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Joyce had business cards made through a printing company. They charge an initial fee of $12 and $3.99 for each box of 100 busine
saw5 [17]

Answer:

$31.95

Step-by-step explanation:

The total cost is the initial fee plus the cost of the boxes.

12 + 5×3.99 = 31.95

4 0
2 years ago
7x+1.1=-3.8 ffffffffffffffffff
emmainna [20.7K]

Answer:

x=-0.7

Step-by-step explanation:

7x+1.1=-3.8

subtract 1.1 on both sides

7x=-4.9

divide both sides by 7

x=-0.7

4 0
2 years ago
What is the slope of each side of the triangle? the sides are A(-2,4) B(-1,1) C(2,3)
Rashid [163]

Answer:

The slopes of three sides of triangle are as follows:

AB = -3

BC = 2/3

AC = -1/4

Step-by-step explanation:

The slope is denoted by m and is calculated using the formula

m = \frac{y_2-y_1}{x_2-x_1}

The given vertices are:

A(-2,4) B(-1,1) C(2,3)

The sides will be:

AB, BC, AC

Let m1 be the slope of AB

Let m2 be the slope of BC

Let m3 be the slope of AC

Now

Slope\ of\ AB = m_1 = \frac{1-4}{-1+2} = \frac{-3}{1} = -3\\Slope\ of\ BC = m_2 = \frac{3-1}{2+1} = \frac{2}{3}\\Slope\ of\ AC = m_2 = \frac{3-4}{2+2} = \frac{-1}{4} = -\frac{1}{4}

Hence,

The slopes of three sides of triangle are as follows:

AB = -3

BC = 2/3

AC = -1/4

6 0
2 years ago
The equation (x-6)^2/16 + (y+7)^2/4 = 1 represents an ellipse. What are the vertices of the ellipse?
topjm [15]

Answer:

The correct option is C.

Step-by-step explanation:

The given equation is

\frac{(x-6)^2}{16}+\frac{(y+7)^2}{4}= 1

It can be rewritten as

\frac{(x-6)^2}{4^2}+\frac{(y+7)^2}{2^2}= 1         .....(1)

The standard form of an ellipse is

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}= 1           ....(2)

Where (h,k) is center of the ellipse.

If a>b, then the vertices of the ellipse are (h\pm a, k).

From (1) and (2) we get

h=6,k=-7,a=4,b=2

Since a>b, therefore the vertices of the ellipse are

(h+a, k)=(6+4,-7)\Rightarrow (10,-7)

(h-a, k)=(6-4,-7)\Rightarrow (2,-7)

The vertices of the given ellipse are (10, –7) and (2, –7). Therefore the correct option is C.

5 0
2 years ago
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2y^4 and 5y^4 because same variable (y), same degree (3)
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3 years ago
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