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3 years ago

12

Really need help. Answer following questions

1 answer:

Shalnov [3]3 years ago

8 0

1. Yes x^2 + 4x + 4 - 4 + 3 = 0 is still equivalent to x^2 + 4x + 3 = 0 since whilst we have added a 4 to the function, we have also subtracted a 4, and since +4 - 4 = 0, we haven't actually changed the value of the equation

2. -1 is the sum of -4 and 3

3. (x + 2)^2 - 1 = (x + 2 + sq.root of 1)(x + 2 - sq.root of 1)
= (x + 2 + 1)(x + 2 - 1)
= (x + 3)(x + 1)
(x + 3)(x + 1) = 0
x = -3 or x = -1

Hope that makes sense :)

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Kevin scores on the first for science test are 88, 92, 82 and 94. What a score must he earn on the fifth test to an average of 9

Debora [2.8K]

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3 years ago

The point $(p,9)$ lies on line $k$ shown here. What is the value of $p$?

blagie [28]

Answer:

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Step-by-step explanation:

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3 years ago

What is the result ?

yuradex [85]

Answer:

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or

Step-by-step explanation:

there are 2 ways

3 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST TO BEST ANSWER- DUE TODAY

solong [7]

Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , with a = x^2 coefficient, b = x coefficient, and c = constant.

Firstly, using our equation plug in the values into our equation (a = 1, b = -4, and c = 3):

$x=\frac{4\pm \sqrt{(-4)^2-4*1*3}}{2*1}$

Next, solve the exponent and multiplications:

$x=\frac{4\pm \sqrt{16-12}}{2}$

Next, solve the subtraction:

$x=\frac{4\pm \sqrt{4}}{2}$

Next, solve the square root:

$x=\frac{4\pm 2}{2}$

Next, you are going to solve this equation twice - once with the + symbol, once with the - symbol as such:

$x=\frac{4+2}{2}\\\\x=\frac{6}{2}\\\\x=3\\\\\\x=\frac{4-2}{2}\\\\x=\frac{2}{2}\\\\x=1$

<u>Your zeros (x-intercepts) are 3 and 1. These solutions are real and rational.</u>

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4 years ago

^O^ HELP PLZZZZ 10P due today !!! Ty <br> ~O~

DENIUS [597]

2 because 6+^82)+=4 hope this help

5 0

3 years ago