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3 years ago

Find the value of x in the following system of equations:

Mathematics

1 answer:

WITCHER [35]3 years ago

3 0

Answer:

c.) x = 3

Step-by-step explanation:

2x - 3y = 9

x = 4 + y

Substitute the second equation into the first equation

2(4+y) - 3y = 9

Distribute

8+2y -3y =9

8-y = 9

Subtract 8 from each side

8-y-8 =9-8

-y = 1

Multiply by -1

y =-1

Now solve for x

x = 4-1

x =3

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3 years ago

Demi’s last 5 bowling scores were 68, 75, 90, 80. What was Femi’s mean score? A.72 B.75 C.77 D.79 D.

-Dominant- [34]

Complete Question:

Femi's last 5 bowling scores were 68, 75, 72, 90, and 80. What was Femi's mean score?

Answer:

Step-by-step explanation:

Given

Scores: 68, 75, 72, 90, and 80

Required

Determine the mean

The mean (M) of a data is calculated as thus:

$M = \frac{\sum x}{n}$

Where

n = number of data.

In this case,

n = 5

and

$\sum x$ = sum of data

So, the formula becomes

$M = \frac{68 + 75 + 72 + 90 + 80}{5}$

$M = \frac{385}{5}$

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Hence, the mean is 77

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3 years ago

The equation a<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class=

andrezito [222]

Answer:

$\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)$

Step-by-step explanation:

We are given the equation:

$ax^2+bx+c=0$

Which has roots α and β.

And we want to express (α + 1)(β + 1) in terms of a, b, and c.

From the quadratic formula, we know that the two solutions to our equation are:

$\displaystyle x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}\text{ and } x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Let x₁ = α and x₂ = β. Substitute:

$\displaystyle \left(\frac{-b+\sqrt{b^2-4ac}}{2a} + 1\right) \left(\frac{-b-\sqrt{b^2-4ac}}{2a}+1\right)$

Combine fractions:

$\displaystyle =\left(\frac{-b+2a+\sqrt{b^2-4ac}}{2a} \right) \left(\frac{-b+2a-\sqrt{b^2-4ac}}{2a}\right)$

Rewrite:

$\displaystyle = \frac{\left(-b+2a+\sqrt{b^2-4ac}\right)\left(-b+2a-\sqrt{b^2-4ac}\right)}{(2a)(2a)}$

Multiply and group:

$\displaystyle = \frac{((-b+2a)+\sqrt{b^2-4ac})((-b+2a)-\sqrt{b^2-4ac})}{4a^2}$

Difference of two squares:

$\displaystyle = \frac{\overbrace{(-b+2a)^2 - (\sqrt{b^2-4ac})^2}^{(x+y)(x-y)=x^2-y^2}}{4a^2}$

Expand and simplify:

$\displaystyle = \frac{(b^2-4ab+4a^2)-(b^2-4ac)}{4a^2}$

Distribute:

$\displaystyle = \frac{(b^2-4ab+4a^2)+(-b^2+4ac)}{4a^2}$

Cancel like terms:

$\displaystyle = \frac{4a^2+4ac-4ab}{4a^2}$

Factor:

$\displaystyle =\frac{4a(a+c-b)}{4a(a)}$

Cancel. Hence:

$\displaystyle = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)$

Therefore:

$\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}$

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3 years ago