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3 years ago

12

Multiply the following polynomials, then place the answer in the proper location on the grid. Write the answer in descending pow

ers of x.
(8x + 3)(4x + 7)

1 answer:

attashe74 [19]3 years ago

4 0

32x^2+64x+12x+21 => 32x^2+76x+21 you multiply the two equation together and you will get this

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James says that 12 is a common factor of 3 and 4.june says that 12 is common multiple of 3 and 4.who is correct.

Sliva [168]

June is correct because <span>12 is a </span>multiple<span> of both </span>3 and 4<span>.</span>

7 0

3 years ago

Read 2 more answers

The figure shows the dimensions of a room in which receptions are held. the room is being carpeted. The three semi-circular part

Mice21 [21]

A=LW so the 18m*12m= 216 m^2

4 0

3 years ago

In ΔABC, which trigonometric ratio has the value a/c

valentinak56 [21]

Answer:

$\tan A$

Step-by-step explanation:

From the diagram, side length $a$ is opposite to angle A and side length $c$ is adjacent to angle A.

Recall the mnemonics SOH CAH TOA.

We use TOA, because it involves opposite and adjacent.

$\tan A=\frac{Opposite}{Adjacent}$

$\tan A=\frac{a}{c}$

The correct choice is A.

7 0

4 years ago

Read 2 more answers

If p(x) = x 3 + x 2+ root 5 x + root 5 , then find p( – 5 )

Morgarella [4.7K]

Answer:

-100-4√5

Step-by-step explanation:

Given the polynomial expression

p(x) = x³+x²+√5x + √5

p(-5) = (-5)³+(-5)² + √5(-5) + √5

p(-5) = -125 + 25 -5√5 + √5

p(-5) = -100-4√5

This gives the required answer

6 0

3 years ago

How many 3-letter sequences can be formed if the second letter must be a vowel (A, E, I, O, or U), and the third letter must be

erica [24]

Answer:

120 ways

Step-by-step explanation:

Given that, 3-letter sequences should be formed, where the second letter must be a vowel (A, E, I, O, U) and the third letter must differ from the first letter.

There are 26 alphabets in English,

so, firstly let us fix the second letter to be one of the vowels

now we are left with 25 alphabets and now let us fix the third letter to be an alphabet from the remaining 25 alphabets.

now, we are left with 24 alphabets and the first letter can be any of these 24 alphabets.

so, there are 24 ways to fill the sequence.

The same thing can be done with the other 4 vowels.

so, total number of ways = 5 × 24 = 120 ways.

8 0

3 years ago