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anastassius [24]
3 years ago
5

Find the area of the Compound shape The answer is 333 cm^2 I just need the steps....

Mathematics
1 answer:
Serggg [28]3 years ago
8 0

Answer:

333cm^2

Step-by-step explanation:

Divide the compound shape into the two separate parts, the rectangle and the triangle.

The rectangle measuring 18cm by 14cm, getting the area, use the formula length*width. This is 18*14= 252cm^2.

Then the triangle has the height, 9cm and base 18cm, finding the area use the formula, 1/2*base*height.

1/2*9*18= 81cm^2.

Add the two, 252+81= 333cm^2.

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WILL GIVE BRAINLEST SUPER EASY A coyote can run up to 43 miles per hour while a rabbit can run up to 35 per hour. Write two equi
lakkis [162]

<u>48</u><u> </u><u>more</u><u> </u><u>miles</u>

Coyote= 43h

Rabbit= 35h

Coyote= 43(6)= 258

Rabbit= 35(6)= 210

Now subtract 258 by 210

258-210= <u>48</u>

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3 years ago
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Molodets [167]
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5 0
3 years ago
What is the answer to 284 times 45 to the power of 3
Alecsey [184]

Hey there!

284 * 45^3

= 284 * 45 * 45 * 45

= 284 * 2,025 * 45

= 284 * 91,125

= 25,879,500


Therefore, your answer is:

25,879,500



Good luck on your assignment & enjoy your day!


~Amphitrite1040:)

5 0
2 years ago
This is the first step in which construction?
slavikrds [6]

Answer:

Option A is correct.

Step-by-step explanation:

The first step in the construction of a square inscribed in a circle is to draw a diameter of the circle.

Then we have to draw perpendicular bisector of the diameter using a compass.

Therefore, the perpendicular bisector will be another diameter of the circle and both the diameters will meet at the center of the circle.

Now, join the points with straight lines in order where the two diameters meet the circle. And finally, we will get a square inscribed in the circle.

Hence, option A is correct. (Answer)

5 0
3 years ago
Part C: Find the distance from B to E and from P to E. Show your work. (4 points)
garri49 [273]

Answer:

BE = 294.23 ft

PE = 259.62 ft

Step-by-step explanation:

Quadrilateral CPRG is a parallelogram.

CP || GR

In triangles BCG and BPE

\angle BCG \cong \angle BPE  (Alternate \: \angle 's) \\\angle GBC \cong \angle EBP  (Vertical \: \angle 's) \\\therefore \triangle BCG \sim \triangle BPE  (AA \: postulate) \\\\\therefore \frac{BE}{BG}  =\frac{PE}{GC} =\frac{BP}{BC}...(csst)\\\\\therefore \frac{BE}{425}  =\frac{PE}{375} =\frac{225}{325}\\\\\therefore \frac{BE}{425}  =\frac{PE}{375} =\frac{9}{13}\\\\\therefore \frac{BE}{425} =\frac{9}{13}\\\\BE =\frac{9\times 425}{13}\\\\BE =\frac{3,825}{13}\\\\BE = 294.23\: ft \\\\\&\\\frac{PE}{375} =\frac{9}{13}\\\\PE =\frac{9\times 325}{13}\\\\PE =\frac{3,375}{13}\\\\PE = 259.615385 \\\\PE = 259.62 \: ft\\\\

7 0
3 years ago
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