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faltersainse [42]

3 years ago

6

Please help! What is the product in the form ax^2+bx+c

1 answer:

Firlakuza [10]3 years ago

3 0

Answer:

a=2, b=2, c=-12

Step-by-step explanation:

Since it's in factor form, you would first distribute using foil to get $2x^{2} -4x+6x-12$ and then combine like terms when necessary. In this case, the fully simplified version is $2x^{2} +2x-12$ . Therefore your a is 2, b is 2, and c is -12.

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Which of the following equations is equivalent to the logarithmic equation<br> below?<br> x = In4

Ne4ueva [31]

Given:

The equation is:

$x=\ln 4$

To find:

The equations that is equivalent to the given logarithmic equation.

Solution:

We have,

$x=\ln 4$

It can be written as:

$e^x=e^{\ln 4}$

$e^x=4$ $[\because e^{\ln x}=x]$

Therefore, the correct option is D.

7 0

3 years ago

Prove the sum of two rational numbers is rational where a, b, c, and d are integers and b and d cannot be zero. Steps Reasons 1.

harkovskaia [24]

Answer:

$\dfrac{ad+bc}{bd}$

Step-by-step explanation:

Let $\frac{a}{b}$ and $\frac{c}{d}$ be two rational numbers, where b and d are not zero and a, b, c and d are integers.

1. Given:

$\dfrac{a}{b}+\dfrac{c}{d}$

2. Multiply to get a common denominator :

$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad}{bd}+\dfrac{cb}{db}$

3. Simplify:

$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad}{bd}+\dfrac{cb}{db}=\dfrac{ad+bc}{bd}$

4. Since $b\neq 0,\ d\neq 0,$ then $bd\neq 0.$

If $a,b,c,d$ are integers, then $bd, ad,bc, ad+bc$ are integers too. So the fraction

$\dfrac{ad+bc}{bd}$

is a rational number

6 0

3 years ago

Read 2 more answers

What matches with Pictograph??

nika2105 [10]

Answer:

8

Step-by-step explanation:

for brainliest

4 0

2 years ago

Read 2 more answers

What is the area of a circle with a diameter of 26 centimeters?

Gennadij [26K]

A=πr²

A=π13²

A=530.93
hope this helped

8 0

3 years ago

Please help me on this problem.

VashaNatasha [74]

Answer : (8,5)

Given

K(x) = $8-\sqrt[3]{x}$

Now we plug in each point and check whether it satisfies our equation

(-64,12)

K(-64) = $8-\sqrt[3]{-64}$ = 12 so this point lie on graph

(125,3)

K(125) = $8-\sqrt[3]{125}$ = 3 so this point lie on graph

(343,1)

K(343) = $8-\sqrt[3]{343}$ = 1 so this point lie on graph

(8,5)

K(8) = $8-\sqrt[3]{8}$ = 6 so this point does not lie on graph because we got 6 instead of 5

4 0

3 years ago