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3 years ago

What are the solutions of the quadratic equation 4x^2-8x-12=0

Mathematics

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

Step-by-step explanation:

$4x^2-8x-12=0\\{}\qquad\qquad^{\div4\qquad\div4}\\x^2-2x-3=0\\\\x^2+x-3x-3=0\\\\x(x+1)-3(x+1)=0\\\\(x+1)(x-3)=0\\\\x+1=0\quad\vee\quad x-3=0\\\\x=-1\quad\vee\quad x=3$

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Exponential form of log4 4=1

vlabodo [156]

4^1= 4. the exponential form is b^e=n, while log is logb n=1.

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3 years ago

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wariber [46]

I agree with the 14x all right, but the remainder of the other part is 178. I'm going to recheck my answer. I'm not confident about using synthetic division for this question. It is 57 years since I did something like this. If I was more confident, I would list the results of synthetic division because it is much easier to show than long division. I'm saying the answer should be 14x + 178.

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3 years ago

Read 2 more answers

Bread and sugar cost 110 together,bread cost 100 more than sugar,how much does sugar cost?

dangina [55]

First write an equation system based on the problem
We can write "<span>Bread and sugar cost 110 together" as
</span>∴ b + s = 110
We can write "<span>bread cost 100 more than sugar" as
</span>∴ b = 100 + s
<span>
Second, solve the equation system by subtitution method
Subtitute b with (100+s) in the first equation, and we'll find the value of s
b + s = 110
(100 + s) + s = 110
100 + 2s = 110
2s = 110 - 100
2s = 10
s = 10/2
s = 5
The cost of the sugar is 5</span>

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3 years ago

Write an equation in slope-intercept form which has a slope of 1/4 and crosses the Ÿ-axis at (0,7)

12345 [234]

Answer:

y= 1/4 x +7

Step-by-step explanation:

The slope intercept form of a line is

y = mx + b where m is the slope and b is the y intercept

the slope is 1/4 and the y intercept is 7

y= 1/4 x +7

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2 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$