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2 years ago

One question WILL GIVE BRAINLIEST - Algebra

Mathematics

1 answer:

snow_tiger [21]2 years ago

6 0

Answer:

B' (5,1)

C' (3,2)

A' (3,0)

Step-by-step explanation:

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Solve<br><img src="https://tex.z-dn.net/?f=%7B2x%7D%5E%7B2%7D%20%20%2B%208x%20-%201%20%3D%200" id="TexFormula1" title="{2x}^{2}

kramer

Answer:

x= ± $3\sqrt{2} /2$ +2

Step-by-step explanation:

Use the formula ( b/2) ^2 in order to create a new term. Solve for x by using this term to complete the square.

x= ± $3\sqrt{2} /2$ +2

4 0

7 months ago

Which number line model represents the expression -2+(-7)?<br> Khan academy help!!!!!

gtnhenbr [62]

-9 because 2+7=9 so the negative part oh then must be -9

6 0

2 years ago

kotykmax [81]

Answer:

16.5 hours

Step-by-step explanation:

The bicycle costs $240. Eva already has $42. She is making $12 per hour.

240 = 12x + 42

Solve for x to find out how many hours Eva needs to work to buy the bicycle.

240 = 12x + 42

240 - 42 = (12x + 42) - 42

198 = 12x

198/12 = (12x)/12

16.5 = x

Eva needs to work 16.5 hours to be able to buy the bikw.

6 0

3 years ago

8. Let X be the number of cars per minute passing a certain point of some road between 8am and 10am on a Sunday. Assume that X h

Lana71 [14]

Answer:

P (3 or fewer) =0.2650

Step-by-step explanation:

Mean = x` = 5

The Poisson distribution formula is given by

P(X) = e-ˣ` x`ˣ/ x!

The mean is 5 and the X takes the values 0,1,2,and 3 which means 3 or fewer, so we add the probability of all the values of X to get the desired Value of X.

P(3 or fewer ) = e-⁵ (5)³/3!+ e-⁵ (5)²/2! +e-⁵ (5)/1!+e-⁵ (5)⁰/0!

Putting the Values

P (3 or fewer) = 0.006737 . 125 / 6 + 0.006737 . 25 / 2 +0.006737 . 5 / 1 + 0.006737 . 1 / 1

P (3 or fewer) = 0.140374 + 0.08422+ 0.03369 +0.006737

P (3 or fewer) =0.2650

7 0

3 years ago

How many trucks carried only late variety?

murzikaleks [220]

9514 1404 393

Answer:

late only: 15
extra-late only: 24
one type: 43
total trucks: 105

Step-by-step explanation:

It works well when making a Venn diagram to start in the middle (6 carried all three), then work out.

For example, if 10 carried early and extra-late, then only 10-6 = 4 of those trucks carried just early and extra-late.

Similarly, if 30 carried early and late, and 4 more carried only early and extra-late, then 38-30-4 = 4 carried only early. In the attached, the "only" numbers for a single type are circled, to differentiate them from the "total" numbers for that type.

a) 15 trucks carried only late

b) 24 trucks carried only extra late

c) 4+15+24 = 43 trucks carried only one type

d) 38+67+56 -30-28-10 +6 +6 = 105 trucks in all went out

8 0

2 years ago