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dmitriy555 [2]
1 year ago
6

20 cm

Mathematics
1 answer:
lawyer [7]1 year ago
7 0

The area of the arrow given in the figure is 610 square cm

<h3>Area of composite figure</h3>

The given figure is made up of rectangle and triangle. The area is expressed as:

Area = Area of rectangle + area of triangle

Substitute the given parameters

Area of the arrow = (15*20) + 0.5(31 * 20)

Area of the arrow = 300 + 310
Area of the arrow = 610 square cm

Hence the area of the arrow given in the figure is 610 square cm

Learn more on area of composite figures here: brainly.com/question/21135654

#SPJ1

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The equation of a line is y = 1.5x − 2. What are its slope and y-intercept?
andre [41]

This equation is written in slope intercept form, y=mx+b, with m being the slope and b being the y-intercept.

The slope of the line is 1.5 and the y-intercept is -2. The answer is A.

I hope this helps ;)

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3 years ago
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Let a and b be real numbers where ab+0. Which of the following functions could represent the graph below?
8090 [49]

Answer:

O f(x) = x(x – b) 3

Step-by-step explanation:

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2 years ago
Alice plants carrots in a rectangular garden bed. The garden bed has a width of 1.5 meters
mojhsa [17]

Answer:

1. L= A/W

2. 4m

Step-by-step explanation:

We can define this problem as a word problem relating to mensuration of flat shapes, a

rectangle

1. Let the length be L and the width be W

Given

Area A= 6m²

W= 1.5m

We know that the area of a rectangular shape is

Area = L x W

The expression for the length is

L= A/W

2. The solution for the length is

6=1.5L

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8 0
3 years ago
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Solve the following ODE's: c) y* - 9y' + 18y = t^2
Nastasia [14]

Answer:

y = C_1e^{3t}+C_2e^{6t} + \dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = C_1e^{3t}+C_2e^{6t}

now calculating P.I.

P.I. = \frac{t^2}{D^2 - 9D +18}

P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})

hence the complete solution

y = C.F. + P.I.

y = C_1e^{3t}+C_2e^{6t} + \dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})

7 0
2 years ago
Please help ASAP<br><br>i dont understand how to do this.
allochka39001 [22]

Circumference of a circle=2πr

Where r is radius

C=(2*2.2)π

C=4.4π



8 0
3 years ago
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