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nikklg [1K]
4 years ago
7

Write the equation of a line perpendicular to 8xplus5yequals2 that passes through the point ​(2​,negative 7​).

Mathematics
1 answer:
SSSSS [86.1K]4 years ago
7 0

8x + 5y = 2   →   5y = - 8x + 2   →   y = (-8/5)x + 2  ⇒ m = -8/5

perpendicular slope = 5/8

Point-Slope formula: y - y₁ = m(x - x₁)  ; where (x₁, y₁) is the given point

y - (-7) = (5/8)(x - 2)

y + 7 = (5/8)(x - 2)

8(y + 7) = 5(x - 2)

8y + 56 = 5x - 10

8y + 66 = 5x

        66 = 5x - 8y   ⇒   5x - 8y = 66

Answer: 5x - 8y = 66


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The circle shown below has a diameter of 18 centimeters. What is the area of the shaded sector?
N76 [4]
Area of circle = Pi x r*2

If diameter is 18, radius is 9.

Area = pi (3.14) x r*2 (9*2 = 81) .. 3.14 x 81 = 254.34 sq cen = total area of circle.

A circle is 360 degrees, the shaded area is 270 degrees. the shaded area is 3/4 (270/360) of the area of the circle.

254.34 x 3/4 = 190.755 sq cen

Hope this helps
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3 years ago
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Charra [1.4K]
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 20 boxes -----> 15 min
100 boxes ----> x
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x = 75
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7 0
3 years ago
Given: KLMN is a trapezoid m∠K = 90°, m∠N = 45° LK = LM = 10 Find: KN, Area of KLMN
Ilya [14]
Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to A= \frac{LM+KN}{2} *MP= \frac{10+20}{2} *10=150 sq. units.


5 0
3 years ago
A triangular prism is shown.<br> What is the surface area of the triangular<br> prism?
Brilliant_brown [7]

Answer:

Area= 840

Step-by-step explanation:

Area= (20x13)x2+(10x12x1/2)x2+10x20

=520+120+200

=840

  • Area=Two area of the triangle [13x20] + Two area of the rectangle + Under area (rectangle)- [10x20]
5 0
3 years ago
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