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Leya [2.2K]
3 years ago
14

If the length of a rectangle is x + 4 and the width is 2x – 3, what is the area?

Mathematics
1 answer:
zmey [24]3 years ago
6 0
The area of a rectangle is the length times the width.

(x+4) \times (2x-3)=2x^2-3x+8x-12=2x^2+5x-12

The answer is C.
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PLEASE HELP ME!! What is 3,200 as a product of its factors?? This for 6th grade
lakkis [162]

Answer:

3,200 is first divided by 2 which is smallest prime number . Then the answer is get 1,600 . Again this number is divided by 2 then answer is 800 . This step is continues for all time . So , it is again divided by 2 then answer comes 400 then divided answer comes 200. Then again divided by 2 then answer is 100 and again divided by 2 then answer is 50 again divided by 2 then answer is 25. now we cannot divided by 2 this number 50. Then what to do? we have to use prime number .so we have to divided by 5 then answer is 5. In last ,for checking we have to multiply all this prime number to check this answer.

7 0
3 years ago
3(x-4)=12x distributing or diving
frutty [35]
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7 0
3 years ago
What is AE ? Enter your answer in the box. units Two segments A D and B C intersect at point E to form two triangles A B E and D
spin [16.1K]

Answer:

The length of AE is 20 units.

Step-by-step explanation:

Given two segments AD and BC intersect at point E to form two triangles ABE and DCE. Side AB is parallel to side DC. A E is labeled 2x+10. ED is labeled x+3. AB is 10 units long and DC is 4 units long.

we have to find the length of AE

AB||CD ⇒ ∠EAB=∠EDC and ∠EBA=∠ECD  

In ΔABE and ΔDCE

∠EAB=∠EDC      (∵Alternate angles)

∠EBA=∠ECD      (∵Alternate angles)

By AA similarity, ΔABE ≈ ΔDCE

therefore, \frac{AE}{ED}=\frac{AB}{CD}

⇒ \frac{2x+10}{x+3}=\frac{10}{4}

⇒ 8x+40=10x+30

⇒ x=5

Hence, AE=2x+10=2(5)+10=20 units

The length of AE is 20 units.

5 0
3 years ago
(x + 7)(x + 10) = what is the answer
ch4aika [34]

We can use FOIL to solve.

(x + 7)(x + 10)

(x * x) + (x * 7) + (7 * x) + (7 * 10)

x² + 7x + 7x + 70

x² + 14x + 70

Best of Luck!

8 0
3 years ago
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