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3 years ago

What is the area of the figure below?

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

6 0

A=(a+b/2) x h

a = 8
b = 14
h = 14

(8+14/2) x 14

(22/2) x 14

11 x 14 = 154

The answer is 154

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The sum of two numbers is 50 and the difference is 10. What are the numbers?

Black_prince [1.1K]

Answer:

I believe the answer is 20 and 30

5 0

3 years ago

Cows and horses cost $250 each and sheep and goats each cost $200 at the county fair. Chickens sell for $50. How much could McDo

IgorC [24]

Answer:

the total amount of animals in McDonald's farm is missing, so I looked for a similar question and found the attached image:

McDonald can earn:

15 cows x $250 = $3,750

2 horses x $250 = $500

20 sheep x $200 = $4,000

25 goats x $200 = $5,000

17 chickens x $50 = $850

total = $14,100

6 0

3 years ago

Which of these choices is the length of the leg of a 45-45-90 triangle with a hypotenuse of 20?

Nuetrik [128]

The answer would be b.20

7 0

3 years ago

Read 2 more answers

Select the polynomial that is a perfect square trinomial.

alexdok [17]

Answer:

Option D. It's a perfect square trinomial.

Step-by-step explanation:

(a) 36x² - 4x + 16

= (6x)² - 2(2x) + (4)²

It's not a perfect square trinomial

(b) 16x² - 8x + 36

= (4x)² - 2x(4x) + (6)²

It's not a perfect square trinomial

= (5x)² + 2 $(\frac{9}{2}x)$ + (2)²

It's not a perfect square trinomial

(d) 4x² + 20x + 25

= (2x)² + 2(10x) + (5)²

= (2x+5)²

It's a perfect square trinomial.

8 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago