<img src="https://tex.z-dn.net/?f=3x%20%5E%7B2%7D%20%2B%2010x%20%2B%208" id="TexFormula1" title="3x ^{2} + 10x + 8" alt="3x ^{2}

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olganol [36]

3 years ago

+ 10x + 8" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

8 0

if we're factoring you can use the AC method. A multiplied by C is 24. and 4x6 is 24.

(x+4)(x+6) is your answer

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The equation of a line is 3x−5y=8. Select three lines that are parallel to the given line.

sweet-ann [11.9K]

Answer:

The Given Equation is

3x - 5y = 8

Slopes of paralell lines are equal

So option 3, 4, 5 are correct

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Hope it helps..

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3 years ago

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3 years ago

Quadratic Formula to solve the equation x^2- 4x = - 7

mylen [45]

roots of the equation x^2- 4x = - 7 are 2+3i and 2-3i

What are the natures of root of quadratic equation?

Case I: $4ac > b^2$

The roots of the quadratic equation $ax^2 + bx + c = 0$ are real and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive.

Situation II: $b^2$ $-4ac$ =0

The roots and of the quadratic equation $ax^2 + bx + c = 0$ are real and equal when a, b, and c are real numbers, a 0, and the discriminant is zero.

Case III: $b^2$ $-4ac$ <0

The quadratic equation $ax^2 + bx + c = 0$ has roots when a, b, and c are real numbers, a 0, and the discriminant is negative, but these roots are not equal and are not real. We refer to the roots in this instance as fictitious.

Case IV: Perfect square and $b^2 - 4ac > 0$

The roots of the quadratic equation $ax^2 + bx + c = 0$ are real, rational, and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive and perfect square.

Quadratic Formula $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

$x^2- 4x = - 7$

$x^2-4x+7=0$

Here a = 1 , b = -4 , c = 7

using quadratic formula

$x={\frac {-(-4)\pm {\sqrt {(-4)^{2}-4(1)(7)}}}{2(1)}}$

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

$x={\ {2\pm {\ 3i}}$

Learn more about quadratic equation from the link below

brainly.com/question/2279540

#SPJ1

8 0

1 year ago

On a given day, machine A has a 10% chance of malfunctioning and machine B has a 7% chance of the same. Given that at least one

valentinak56 [21]

I will give an instance to answer this. Suppose that we have 1,000 days. And it was given above that machine A has 10 % chances to malfunctioned and B has a 7%. So machine A has 100 days that malfunctioned while machine B has 70 days. And 7 days that both machines malfunctioned (10%x7%). Next is to add 100 days from machine A and 70 days from machine B and subtract with the 7 days where both of them malfunctioned. So the result is 163 days out of 1000 that we expect an incidence of failure. So the chance of machine B to malfunctioned is 70 days. 70/163 is 42.93%

4 0

3 years ago