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3 years ago

7

Please help me. right answer will get brainliest

1 answer:

Harman [31]3 years ago

8 0

Answer:

I attached the answers :)

Step-by-step explanation:

hope this helps :)

(Its a bit messy becuz i used a mouse to write, sry about that)

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Plzzz I need help with this question I tried to solve it many times but I can't

blsea [12.9K]

Answer and Step-by-step explanation: Area of a right triangle, (as any other triangle), is calculated as: $A1=\frac{(base)(height)}{2}$

Area of a rectangle is calculated as: $A2=(side)(side)$

Area of a right trapezoid is: $A3=\frac{(a+b)h}{2}$ , where:

a is short base

b is long base

h is height

1) Expressing areas in terms of x:

Area of triangle S1:

$S1=\frac{(2x-3)(4x-6)}{2}$

$S1=4x^{2}-12x+9$

Area of rectangle S2:

$S2 = (4x-6)(3x-2)$

$S2=12x^{2}-26x+12$

Area of trapezoid S3:

$S3=\frac{(2x+3+4x+1)(2x-3)}{2}$

$S3=\frac{(6x+4)(2x-3)}{2}$

$S3=6x^{2}-5x-6$

2) a) $S=4x^{2}-12x+9+12x^{2}-26x+12-(6x^{2}-5x-6)$

$S=4x^{2}-12x+9+12x^{2}-26x+12-6x^{2}+5x+6$

$S=10x^{2}-33x+37$

Which is the same as S = (2x-3)(5x-9)

b) For the areas to be the same:

$\frac{(3x-2+3x-2+2x-3)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$\frac{(8x-7)(4x-6)}{2}=\frac{(6x+4)(2x-3)}{2}$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12$

$20x^{2}-66x+54=0$

Using Bhaskara to solve the second degree equation:

$\frac{66+\sqrt{(-66)^{2}-(4.20.54)} }{2(20)}$

$x_{1}=\frac{66+6}{40}$ = 1.8

$x_{2}=\frac{66-6}{40}$ = 1.5

For the areas of AFGC and ADEB to be equal, x has to be 1.5 or 1.8.

c) <u>Expand</u> <u>a</u> <u>polynomial</u> (or equation) is to multiply all the terms, remiving the parenthesis. <u>Reduce</u> <u>a</u> <u>polynomial</u> (or equation) is to combine terms alike,e.g.:

$S=(2x-3)(5x-9)$

$S=10x^{2}-18x-15x+27$ (expand)

$S=10x^{2}-33x+27$ (reduce)

d) For area of AFCG to be bigger than area of ADEB by 27:

$32x^{2}-48x-28x+42=12x^{2}+8x-18x-12+27$

$32x^{2}-48x-28x+42=12x^{2}+8x-18x+15$

$20x^{2}-66x+27=0$

Solving:

$\frac{66+\sqrt{(-66)^{2}-(4.20.27)} }{2(20)}$

$\frac{66+46.86}{40}$

$x_{1}=\frac{66+46.86}{40}=$ 2.82

$x_{2}=\frac{66-46.86}{40}$ = 0.48

According to the enunciation, x cannot be less than 1.5, then, the value of x so that area AFGC exceeds the area ADEB by 27 is 2.82

6 0

3 years ago

Which expression is equivalent to (3a)-2?<br> O<br> o<br> 3a2<br> ܝ | ܝ | ܚ| ܘ ܩ|ܛܘ

Bad White [126]

Number 1 is correct 1/9a2

6 0

3 years ago

Q2) 7x - 11 = 73<br> Q3) 72 - 6 = 29

almond37 [142]

Answer:

Q2) x=12

Q3) 66=29

Step-by-step explanation:

Q2) 7x−11=73

Step 1: Add 11 to both sides.

7x−11+11=73+11

7x=84

Step 2: Divide both sides by 7.

7x/7=84/7

x=12

Question 3:

Subtract 6 from 72

66 = 29

4 0

3 years ago

When h is one-half and j is one-third, g is 4. if g varies jointly with h and j, what is the value of g when h is 2 and j is 3?

Softa [21]

I don’t know the answer to this question

3 0

3 years ago

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George and his friends rode motorcycles all day on Saturday. He made a table that showed the number of gallons of gas remaining

salantis [7]

Answer:

negative correlation

Step-by-step explanation:

his fuel keeps decreasing every time he uses his motorcycle

5 0

2 years ago