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Tom [10]
2 years ago
11

What should be added to (-5 a2b - 9 ab) to get a2b + 3 ab?​

Mathematics
1 answer:
Doss [256]2 years ago
5 0

Based on given conditions,

a²b + 3 ab - (-5 a²b - 9 ab)

→a²b + 3 ab + 5 a²b + 9 ab

→(a²b + 5 a²b) + (3ab + 9 ab

Now collecting the coefficients and adding them,

(1 + 5)a²b + (3 + 9)ab

→6 {a}^{2}b + 12ab

Hence, the <u>6</u><u>a</u><u>²</u><u>b</u><u> </u><u>+</u><u> </u><u>1</u><u>2</u><u>a</u><u>b</u> should be added.

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Step-by-step explanation:

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Beth's mother can sew 235 pairs of short pants in 6 days while lourdes can sew 187 pairs in 8 days . How many more pants of shor
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Adding whole numbers what is 73404 + 27865
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3 years ago
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I have to use completing the square then put it in vertex form, and I have no clue how the hell do to that.
vodka [1.7K]

Answer:

y=8(x+1)²-2

Step-by-step explanation:

Completing the square is a method of factorising. x²+bx+c=0 becomes x²+bx+(\frac{1}{2}b)²-(\frac{1}{2}b)²+c=0. This then gets converted to make an equation of (x+\frac{1}{2}b)²+(-(\frac{1}{2}b)²+c)=0.

Vertex form is y=a(x-h)²+k.

8x²+16x+6=0 must first be divided by 8 to 8(x²+2x+\frac{6}{8})=0 (as there cannot be a coefficient for x²).

This will become 8(x²+2x+\frac{2}{2}²-\frac{2}{2}²+\frac{3}{4})=0.

From there, using the method above, we can convert it to 8[(x+1)²-1+\frac{3}{4}]=0.

Solving this gets 8[(x+1)²-\frac{1}{4}]=0.

With 8 as a, 1 as -h, and 8×-\frac{1}{4} (-2) as k, we can input this into vertex format to get the result of y=8(x+1)²-2.

**This equation involves completing the square and vertex form, which you may wish to revise. I'm always happy to help!

5 0
3 years ago
7. A car dealership has 8 red, 9 silver, and 5 black cars on the lot. Seven cars are randomly chosen to be displayed(in no parti
MrRissso [65]

Answer:

a) 0.0002 = 0.02% probability that all are silver.

b) 0.0033 = 0.33% probability that 5 are red and 2 are black

c) 0.0345 = 3.45% probability that 4 are red and the rest are silver

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question, the order in which the cars were selected is not important, so we use the combinations formula.

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)!}

(a) all are silver

Desired outcomes:

7 silver cars, from a set of 9. So

D = C_{9,7} = \frac{9!}{7!(9-7)!} = 36

Total outcomes:

7 cars from a set of 8 + 9 + 5 = 22. So

T = C_{22,7} = \frac{22!}{7!(22-7)!} = 170544

Probability:

p = \frac{D}{T} = \frac{36}{170544} = 0.0002

0.0002 = 0.02% probability that all are silver.

(b) 5 are red and 2 are black

Desired outcomes:

5 red, from a set of 8

2 black, from a set of 5

D = C_{8,5}*C_{5,2} = \frac{8!}{5!(8-5)!}*\frac{5!}{2!(5-2)!} = 56*10 = 560

Total outcomes:

7 cars from a set of 8 + 9 + 5 = 22. So

T = C_{22,7} = \frac{22!}{7!(22-7)!} = 170544

Probability:

p = \frac{D}{T} = \frac{560}{170544} = 0.0033

0.0033 = 0.33% probability that 5 are red and 2 are black

c) 4 are red and the rest are silver

Desired outcomes:

4 red, from a set of 8

3 silver, from a set of 9

D = C_{8,4}*C_{9,3} = \frac{8!}{4!(8-4)!}*\frac{9!}{3!(9-3)!} = 70*84 = 5880

Total outcomes:

7 cars from a set of 8 + 9 + 5 = 22. So

T = C_{22,7} = \frac{22!}{7!(22-7)!} = 170544

Probability:

p = \frac{D}{T} = \frac{5880}{170544} = 0.0345

0.0345 = 3.45% probability that 4 are red and the rest are silver

7 0
3 years ago
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