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3 years ago

What is the product of 2x+3 and 4x² - 5x+6?

Mathematics

1 answer:

Pie3 years ago

5 0

Answer:

none

Step-by-step explanation:

(2x+3)(4x^2-5x+6)

factor over

8x^3-10x^2+12x+12x^2-15x+18

combine like terms

8x^3 + 2x^2 - 3x + 18

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On a map,1 in. represents 12 mi. Two cities are 6 1/2 in. on a map. How far apart are the actual cities? A. 78 mi B. 72 1/2 mi C

Dovator [93]

1 inch= 12 miles

Multiply 6 1/2 inches by total number of miles per 1 inch (12).

= 6 1/2 * 12
convert to improper fraction

= 13/2 * 12
multiply numerators

= (13*12)/2
multiply in parentheses

= 156/2
divide

= 78 miles

ANSWER: (A) 78 miles

Hope this helps! :)

3 0

3 years ago

Plz help i will give brainly!

Irina-Kira [14]

Answer:

The answer is c

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Pani says she should get $3 discount on the price of each shirt and a $3 discount on the price of each pair of jeans. Write and

PilotLPTM [1.2K]

Answer:

(S - 3) + (X - 3) = B

Step-by-step explanation:

S = amount of shirts

X = amount of jeans

B = total amount she would pay

(S - 3) + (X - 3) = B

5 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

3 years ago

Please help I beg please oh please I’ll give brainliest to whoever answers

Darya [45]

Answer: The third option is correct

Step-by-step explanation: With increased trials, experimental probability will grow closer to theoretical probability. Blue pens make up 1/2 of the probability pool.

8 0

3 years ago