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2 years ago

Choose the correct simplification of (5a)(6b).

Mathematics

2 answers:

GalinKa [24]2 years ago

4 0

The simplest form would be 30ab

ad-work [718]2 years ago

4 0

Answer:

D. 30ab

Step-by-step explanation:

5x6=30

AxB=ab

BRAINLIEST IF YOU WILL :D

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The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t

Doss [256]

Answer:

First Case:

$\displaystyle p=\frac{5}{2}\text{ and } d=-2$

Second Case:

$\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}$

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

$6p+2, 4p^2-10, \text{ and } 4p+3$

Since this is an arithmetic sequence, each subsequent term is d more than the previous term, where d is our common difference.

Therefore, we can write the second term as;

$4p^2-10=(6p+2)+d$

And, likewise, for the third term:

$4p+3=(6p+2)+2d$

Let's solve for d for each of the equations.

Subtracting in the first equation yields:

$d=4p^2-6p-12$

And for the second equation:

$2d=-2p+1$

To avoid fractions, let's multiply the first equation by 2. Hence:

$2d=8p^2-12p-24$

Therefore:

$8p^2-12p-24=-2p+1$

Simplifying yields:

$8p^2-10p-25=0$

Solve for p. We can factor:

$8p^2+10p-20p-25=0$

Factor:

$2p(4p+5)-5(4p+5)=0$

Grouping:

$(2p-5)(4p+5)=0$

Zero Product Property:

$\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}$

Then, we can use the second equation to solve for d. So:

$2d_1=-2p_1+1$

Substituting:

$\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}$

So, for the first case, p is 5/2 and d is -2.

Likewise, for the second case:

$\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}$

So, for the second case, p is -5/4, and d is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0

1 year ago

which equation has an a-value of –2, a b-value of 1, and a c-value of 3? a. 0 = –2x2 x 3 b. 0 = 2x2 x 3 c. 0 = –2x2 3 d. 0 = 2x2

ehidna [41]

If you would like to find the matching equation, you can do this using the following steps:

ax^2 + bx + c = 0
a = -2
b = 1
c = 3

-2x^2 + x + 3 = 0

The correct result would be a. 0 = -2x^2 + x + 3.

8 0

2 years ago

Read 2 more answers

What is 24x^2y^6-16x^6y^2+4xy^2 divided by 4xy^2?

sertanlavr [38]

24x^2y^6-16x^6y^2 = 8y^4x^4 +4xy^2

8y^4x^4 +4xy^2 = 12y^6x^5 / 4xy^2

12y^6x^5 / 4xy^2 = 3y^4x^4 ... I think that is the answer! :)

6 0

2 years ago

1. What is the graph of the absolute value equation? y=|x|-9

Kruka [31]

The graph of y = |x| - 9 is V shaped with vertex at (0, -9).

The graph of y = |x + 4| is V shaped with vertex at (-4, 0).

The graph of y = |x + 3| - 4 is V shaped with vertex at (-3, -4).

The graph of y = -1/2|x| is V shaped with vertex at (0, 0)

8 0

2 years ago

Read 2 more answers

= which of the following should not be included in a paragraph proof?

yarga [219]

Answer:

there is no choices

Step-by-step explanation:

3 0

2 years ago