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2 years ago

Find the area of the trapezoid.

Mathematics

1 answer:

UkoKoshka [18]2 years ago

4 0

The answer is 68cm^2. Hope this helps!

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PLZ HELP IM REALLY CONFUSED WILL MARK BRAINLIEST

Ne4ueva [31]

Answer:

${1.2}^{n} \times 2.5$

Step-by-step explanation:

it's simple my dear friend.

first, she runs 3 miles per week.

so in

${a}^{(n - 1)} \times b$

b is 3. the initial value.

increasement by 20% means she runs 1.2 times than before.

$\frac{100 + 20}{100} = \frac{120}{100} = \frac{12}{10} = 1.2$

then a=1.2

so we have now:

${1.2}^{(n - 1)} \times 3 \\ = {1.2}^{n} \times \frac{3}{1.2} \\ = {1.2}^{n} \times 2.5$

and we got it!

4 0

2 years ago

I was given a project to do before May 2022. How many months /weeks do I need to work on my project before the deadline?

Roman55 [17]

Months 7
Weeks 30.5
Plz don't procrastinate tho

7 0

2 years ago

What is the x-coordinate of the point shown in the graph? On a coordinate plane, point A is at (negative 5, negative 7).

Alinara [238K]

The x coordinate would be -5

8 0

3 years ago

Read 2 more answers

The length of a rectangle is 6ft longer than its with. If the perimeter is 44 what is the area

ikadub [295]

Answer:

Area= L*W= 13*7 = 91ft^2

Step-by-step explanation:

L= 6+W

Perimeter of the rectangle=2(L+W)=40

2(6+W+W)=40

2(6+2W)=40

12+4W=40

4W=40-12

4W=28

W=28/4

W=7ft

L=6+7=13ft

Area=L*W=13*7=91ft^2

7 0

3 years ago

A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

Lesechka [4]

Check the picture below.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ P(\stackrel{x_1}{-10}~,~\stackrel{y_1}{-2})\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{4-10}{2}~~,~~\cfrac{6-2}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{4}{2} \right)\implies \stackrel{\textit{center}}{(-3~,~2)} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{\textit{center}}{(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})}\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[4-(-3)]^2+[6-2]^2}\implies r=\sqrt{(4+3)^2+(6-2)^2} \\\\\\ r=\sqrt{49+16}\implies r=\sqrt{65} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{2}{ k})\qquad \qquad radius=\stackrel{\sqrt{65}}{ r} \\[2em] [x-(-3)]^2+[y-2]^2=(\sqrt{65})^2\implies (x+3)^2+(y-2)^2=65$

5 0

2 years ago