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3 years ago

PLEASE HELP! AM BEGGING! IF I GET A 80 OR LOWER AM GOING TO FAIL! PLEASE DOUBLE CHECK YOU WORK!!!! :(( I used up all my points

Mathematics

2 answers:

miskamm [114]3 years ago

6 0

Answer:

1. 22 degrees

2. 122 degrees

3. Angle 1 is supplementary to angle 6

Step-by-step explanation:

1. It is supposed to be a right angle which means that 90. So, to find GHK you must do 90 - 68 = 22. I don't understand why it is incorrect but, I can promise you that is the correct answer

2. A straight angle is 180 degrees. So, you would subtract 180 - 58 = 122.

Therefore, your answer for 2 would not be 112 degrees it would in fact be 122 degrees.

3. Evidence in the picture below:

Hoochie [10]3 years ago

5 0

Answer:

question 1 is 22

question 2 is 122

ask your teacher why u got question 1 wrong

Step-by-step explanation:

90-68 is 22

180-58= 122

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Tell whether –4 is a solution of the inequality. —3x + 2 < 15. YES OR NO?

NikAS [45]

Answer:

YES

Step-by-step explanation:

-3 times -4 = 12

12 + 2 = 14

14 < 15

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3 years ago

Please help. What is the product? 3x^5(2x^2+4x+1)

AleksandrR [38]

Answer:

Third option is the correct answer

Step-by-step explanation:

$3 {x}^{5} \bigg( 2 {x}^{2} + 4x + 1 \bigg) \\ \\ = 3 {x}^{5} \times 2 {x}^{2} + 3 {x}^{5} \times 4x + 3 {x}^{5} \times 1 \\ \\ = 3 \times 2 {x}^{5 + 2} + 3 \times 4 {x}^{5 + 1} + (3 \times 1) {x}^{5} \\ \\ \red{ \bold{= 6{x}^{7} + 12{x}^{6} +3 {x}^{5} }}$

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3 years ago

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) Expand a polynomial (or equation) is to multiply all the terms, remiving the parenthesis. Reduce a polynomial (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

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Step-by-step explanation:

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14/24

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3 years ago

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