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3 years ago

Multi step equation2x +3=-7

Mathematics

2 answers:

Marina CMI [18]3 years ago

6 0

Answer:

if you need to find the value of x then I can help

First the equation looks like this:

2x + 3 = -7

first find the value of "x" which will be -5 then the equation will look like this:

2(-5) then multiply 2 and -5 which is equal to -10

this will make the equation look like this:

-10 + 3 = -7

then evaluate

-10 + 3 = -7

-7 = -7

Therefore the value of "x" is -5

kvv77 [185]3 years ago

5 0

Answer:

x = -5

Step-by-step explanation:

Subtract 3 from each side, so it now looks like this: 2x = -10
Divide each side by 2 to cancel out the 2 next to x. It should now look like this: x = -5

I hope this helps!

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2 years ago

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Final exam scores are normally distributed with a mean of 74 and a standard deviation of 6. Approximately, what percentage of fi

notka56 [123]

Answer:

81.86%

Step-by-step explanation:

We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.

First of all we will find z-score using z-score formula.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{68-74}{6}$

$z=\frac{-6}{6}=-1$

Now let us find z-score for 86.

$z=\frac{86-74}{6}$

$z=\frac{12}{6}=2$

Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.

Using normal distribution table we will get,

$P(-1$

$P(2>Z)=.97725$

We will use formula $P(a$ to find the probability to find that a normal variable lies between two values.

Upon substituting our given values in above formula we will get,

$P(-1$

Upon converting 0.81859 to percentage we will get

$0.81859*100=81.859\approx 81.86$

Therefore, 81.86% of final exam score will be between 68 and 86.

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3 years ago

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3 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$