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Alexus [3.1K]
4 years ago
11

Solve the following equation. - 8x=-6x +10

Mathematics
2 answers:
Step2247 [10]4 years ago
3 0

Answer:

x=-5

Step-by-step explanation:

-8x=-6x+10

-8x+6x=10

-2x=10

x=-5

makvit [3.9K]4 years ago
3 0

Answer:

x = -5

Step-by-step explanation:

- 8x = - 6x + 10

- 8x + 6x = 10

- 2x = 10

- x = 10/2

- x = 5

x = - 5

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5 0
3 years ago
Solve for x: 2(x 3)2 − 4 = 0 Round your answer to the nearest hundredth. X = 4. 41, 1. 59 x = 1. 34, 5. 24 x = −1. 34, −5. 24 x
gladu [14]

The two values when the provided quadratic equation is solved for the x are -4.41 and -1.59 to the nearest hundredth.

<h3>What is a quadratic equation?</h3>

A quadratic equation is the equation in which the unknown variable is one and the highest power of the unknown variable is two.

The standard form of the quadratic equation is,

ax^2+bx+c=0

Here, (<em>a,b,c</em>) are the real numbers and <em>x </em>is the variable.

To find the value of x, the following formula is used,

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

The given equation is,

2(x+ 3)^2 - 4 = 0

To solve this equation, we need to apply some mathematical operations over it. Let's start with opening the brackets.

2(x+ 3)^2 - 4 = 0\\2(x^2+6x+9)-4=0\\2x^2+12x+14-4\\2x^2+12x+12=0\\x^2+6x+7=0

On comparing with standard equation we get,

a=1, b=6, c=7

Put this values in the above formula,

x=\dfrac{-(6)\pm\sqrt{(6)^2-4(1)(7)}}{2(1)}\\x=\dfrac{-(6)\pm\sqrt{(6)^2-4(1)(7)}}{2(1)}\\x=-4.41,-1.59

Hence, the two values when the provided quadratic equation is solved for the x are -4.41 and -1.59 to the nearest hundredth.

Learn more about the quadratic equation here;

brainly.com/question/1214333

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3 years ago
What is the standard form of y=(x-4)^2-6
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Answer:

standard form : 2x + y = -2

8 0
3 years ago
a pot of water starts at 68 degrees it is put on stove where it heats at a rate of 4 degrees per minute how long will it take to
Vladimir79 [104]
Y=68+4x
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3 years ago
Does anybody know the answer to this ?​
melisa1 [442]

Answer:

Step-by-step explanation:

5 0
3 years ago
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