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bogdanovich [222]
3 years ago
12

Please help I’m really stuck this is my last attempt

Mathematics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

I THINK IT IS 74 NOT 4

I HOPE THIS HELPS!!!!!

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Will give brainliest, thanks, and 5 stars! LOTS OF POINTS! 50 POINTS!
lubasha [3.4K]

Answer in order:

False

True

False

False

False

False

False

True

True

Step-by-step explanation:

All of the solutions to the equation3x^2 - 12 = 0are x = 12 and x = -12

Solve:

Common factor - 3(x^2-4)=0

Divide both sides

x^2-4=0

Use the quadratic formula

x=\sqrt{4,x}=\sqrt{-4

x = 2, x = -2

Thus, This is false..

-------------------------------------------------------------------------------------------------------------

There are two unique solutions to the equations (x-3)^2 = 16

one variable in matrix = one possible value.

Therefore, x = 7, x = -1

Hence this is True

-------------------------------------------------------------------------------------------------------------

The solutions for the equation 2(x-3)^2-18 = 0 are x = 6 and x = 0

2x^3-18x^2+54x-72=0

x=5.080084

Hence this is false

-------------------------------------------------------------------------------------------------------------

The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7

Add both sides by 8

2(x-5)^2-8+8=0+8

2(x-5)^2=8

(x-5)^2=4

x = 7, x = 3

Hence, this is false

-------------------------------------------------------------------------------------------------------------

The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8

(x+3)^2=17

x=\sqrt{17-3,}=-\sqrt{17}-3

Hence this is false.

-------------------------------------------------------------------------------------------------------------

The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4

(2x-1)^2=9

x=2, x = -1

Hence this is false.

-------------------------------------------------------------------------------------------------------------

The only solution for equation (2x-1)^2-49=0 is x = 4

(2x-1)^2=49

x = 4, x = -3

Hence, this is false

-------------------------------------------------------------------------------------------------------------

The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1

3(x+2)^2=3

(x+2)^2=1

x = -1, x = -3

Hence, this is true

-------------------------------------------------------------------------------------------------------------

The solutions for the equation 5x^2 - 180 = 0 are x = 6 and x  = -6

Add 180 to both sides

5x^2-180+180=0+180

5x^2 = 180

x^2 = 36

x=\sqrt{36},x=-\sqrt{36}

x = 6, x = -6

Hence, this is true.

-------------------------------------------------------------------------------------------------------------

[RevyBreeze]

7 0
3 years ago
Find all rational zeros of P(x)=6x^4-x^3-55x^2+9x+9​
o-na [289]

Answer: x = -3, -1/3, 1/2, 3

Step-by-step explanation:

6 0
3 years ago
Write a fraction with a denominator of 6 that is greater than 2/4
lesya [120]
3/6 because 2/4 is really 1/2
7 0
4 years ago
Solve each of the quadratic equations 3x=0.5x2
aalyn [17]

Answer:

3x = 1   \\ x =  \frac{1}{3}

6 0
3 years ago
Can someone help me out with these?
alexgriva [62]

Answer:

x = 7

x = 16

x = √(89)

Step-by-step explanation:

View Image

<u>Image 1 Use:</u>

tan(angle) = opposite/adjacent

<u>Image 2 Use:</u>

cos(angle) = adjacent/hypotenuse

<u>Image 3 Use:</u>

a² + b² = c²

If you're given an angle and 1 side then you can use sin(), cos(), or tan() to solve for the other missing side.

If you're given two sides then you can use the pathagorean theorem, a² + b² = c², to solve for your missing side.

7 0
3 years ago
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