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arsen [322]
3 years ago
11

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.<br> —3x + 2 &lt; 15. YES OR NO?
NikAS [45]

Answer:

YES

Step-by-step explanation:

-3 times -4 = 12

12 + 2 = 14

14 < 15

7 0
3 years ago
Please help. <br><br>What is the product?<br><br>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} }}

8 0
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) <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
Draw or write to show two ways to use a doubles fact to find 6+7
artcher [175]
If you know 6+6 you know 6+7 because 6+7 is 1 more than 6+6.

6+6=12 while 6+7=13.

Another way to look at it is 7+7=14 so 6+7 is 1 less than it making 6+7=13.
5 0
4 years ago
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Ivanshal [37]

Answer:

14/24 or 7/12 simplified

Step-by-step explanation:

substitute values

15 - 1/2(12)

14/24

7/12

5 0
3 years ago
Read 2 more answers
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