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4 years ago

(-4ab+8a-2) - (-4ab-9)

Mathematics

1 answer:

lawyer [7]4 years ago

6 0

Answer:

-a2 • (2a - b - 3c)

Step-by-step explanation:

Equation at the end of step 1 :

((4ab - 23a2) + 12ac)

————————————————————— • a

Step 2 :

-8a2 + 4ab + 12ac

Simplify —————————————————

Step 3 :

Pulling out like terms :

3.1 Pull out like factors :

-8a2 + 4ab + 12ac = -4a • (2a - b - 3c)

Equation at the end of step 3 :

-a • (2a - b - 3c) • a

Step 4 :

Multiplying exponential expressions :

4.1 a1 multiplied by a1 = a(1 + 1) = a2

Final result :

-a2 • (2a - b - 3c)

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michelle drove 325 miles in 5 hours.if she traveled at a steady speed ,how far did she drive per hour

ASHA 777 [7]

Answer:

65 miles per hour

Step-by-step explanation:

325 divided by 5= 65

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3 years ago

Read 2 more answers

Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must

Studentka2010 [4]

Answer:

a) 40 different hamburgers can be ordered with exactly three extras

b) 20 different regular hamburgers can be ordered with exactly three extras

c) 7 different regular hamburgers can be ordered with at least five extras

Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem:

2 options of hamburger(regular or larger)

6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).

(a) How many different hamburgers can be ordered with exactly three extras?

1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

$C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40$

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

$C_{6,3} = \frac{6!}{3!(6-3)!} = 20$

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

$C_{6,5} = \frac{6!}{5!(6-5)!} = 6$

Six extras:

6 extras, from a set of 6. So

$C_{6,6} = \frac{6!}{6!(6-6)!} = 1$

6 + 1 = 7

7 different regular hamburgers can be ordered with at least five extras

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3 years ago

In this activity, you will predict the probability of an event from the relative frequency of the event. Pablo has some one-, fi

Vadim26 [7]

Answer:

Part A

6/40 = 0.15

Part B

16/40 = 0.4

Part C

10/40 = 0.25

Part D

8/40 = 0.20

Part E

The relative frequency of drawing a five-dollar bill is higher than the other relative frequencies. So, I can predict that Pablo is most likely to have more five-dollar bills than any of the others.

Part F

The relative frequency of drawing a one-dollar bill is lower than the other relative frequencies. So, I can predict that Pablo is most likely to have fewer one-dollar bills than bills of any other denomination.

Part G

It would not be a surprise if Pablo had fewer twenties than ones. The experiment was conducted only 40 times, and the numbers of times one-, ten-, and twenty-dollar bills were drawn are not very far apart. So, the number of twenties could be more or less than the number of ones. The same goes for tens and ones.

If you're on Plato an on slide 20 this Answer is for you:

<em>If Pablo does an experiment 100 times, will the relative frequency be more accurate or less accurate than if he did the experiment 40 times? Why?</em>

Answer: As the number of trials increases, the relative frequency becomes closer to the probability of the event. So, the relative frequency would be more accurate if the experiment were repeated 100 times rather than 40 times.

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3 years ago

The basal energy rate is the daily amount of energy (measured in calories) needed by the body at rest to maintain body temperatu

lara [203]

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4 years ago

What is 153/4 as a whole number

maw [93]

$\frac{4}{3} = \frac{51}{n} \\ = > \frac{3}{4} = \frac{n}{51 } \\ = > n = \frac{3}{4} \times 51 \\ = > n = \frac{153}{4} \\ = > n = 38 \frac{1}{4}$

Hope you could get an idea from here.

Doubt clarification - use comment section.

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2 years ago