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2 years ago

15

Which ordered pair represents a point that is a solution of the equation y = −1/2x and therefore, lies on the graph of this func

tion?
note: time4learning

1 answer:

jasenka [17]2 years ago

5 0

Answer:

is there a graph that needs to go with it?

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30. (14 letters) The process of changing one unit of measure to another.

Georgia [21]

Unit conversion
taxable income

7 0

3 years ago

Read 2 more answers

2.<br><br> Find the slope of the line that passes through the points (4,2) and (4,10)

muminat

Answer:

The slope is unidentified

Step-by-step explanation:

The equation of the line is x=4

4 0

2 years ago

The Louvre is an art museum in Paris. A glass square base pyramid sits in front of the entrance to the museum. If the pyramid is

Sveta_85 [38]

Answer:

lateral area = 2320 m²

Step-by-step explanation:

The question wants us to calculate the lateral area of a square base pyramid. The square base pyramid has a side of 40 meters.The height is 21 meters.

Half of the square base is 40/2 = 20 meters . With the height it forms a right angle triangle. The hypotenuse side is the slant height of the pyramid.

Using Pythagoras's theorem

c² = a² + b²

c² = 20² + 21²

c² = 400 + 441

c² = 841

square root both sides

c = √841

c = 29 meters

The slant height of the pyramid is 29 meters.

The pyramid has four sided triangle. The lateral area is 4 multiply by the area of one triangle.

area of triangle = 1/2 × base × height

base = 40 meters

height = 29 meters

area = 1/2 × 40 × 29

area = 580

area of one triangle = 580 m²

Lateral area = 4(580)

lateral area = 2320 m²

5 0

2 years ago

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

7 0

3 years ago

What is the inverse of the function f(x) = x – 12?

Luda [366]

Answer:a

Step-by-step explanation:

h(x)=x-12

h(x)=4x-48

h(x)=-1(4x-48)

h(x)=-4x+48

h(x)=48-4x

7 0

2 years ago