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2 years ago

Need do find the domain and range

Mathematics

2 answers:

alexandr402 [8]2 years ago

6 0

Answer:

Domain: (-∞,∞)

Range: [-105/8,∞)

Step-by-step explanation: Hope this helps.

Morgarella [4.7K]2 years ago

5 0

Answer:

Domain: (-∞,∞)

Range: [ $-\frac{105}{8} ,$ ∞)

Step-by-step explanation:

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murzikaleks [220]

Answer:

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3 years ago

10x 6x+8 Solve For X

zhannawk [14.2K]

Answer:

I had x to be 2

Step-by-step explanation:

a square has all it side to be equal

so 10x=6x+8

you will group like terms 10x-6x=8

4x=8 you divide both side by the coefficient of x which is four so X=2

7 0

3 years ago

The science club is ordering T-shirts by mail. The shirts cost $12 each and there is a $4 shipping charge. The function C = 4 +

Lisa [10]

The total cost of ordering 10 shirts can be calculated as following:

C = 4 + 12(10)

C = 4 + 120

C = 124

8 0

3 years ago

REPOST!!!!!! 20 POINTS!!!!!!

kkurt [141]

Answer:

Step-by-step explanation:

according to tangent property

we get

AM=BM
AN=CN
BK=CK

Given that

AM=6
BK=4

Thus our equation is

$\rm \displaystyle 2AM + 2AN + 2BK = 34$

substitute the given values

$\rm \displaystyle 2.6+ 2AN+ 2.4 = 34$

simplify multiplication:

$\rm \displaystyle 12+ 2AN+ 8 = 34$

simplify addition:

$\rm \displaystyle 2AN+ 20= 34$

cancel 20 from both sides:

$\rm \displaystyle 2AN=14$

divide both sides by 2:

$\rm \displaystyle AN=7$

AN=CN

hence,CN=7

therefore our answer is A

6 0

2 years ago

A consumer organization estimates that over a 1-year period 20% of cars will need to be repaired once, 8% will need repairs

Sladkaya [172]

Answer:

Probability that a car need to be repaired once = 20% = 0.20

Probability that a car need to be repaired twice = 8% = 0.08

Probability that a car need to be repaired three or more = 2% = 0.02

a) If you own two cars what is the probability that neither will need repair?

Probability that a car need to be repaired once , twice and thrice or more= 0.20+0.08+0.02=0.3

Probability that car need no repair = 1-0.3=0.7

Neither car will need repair= $0.7 \times 0.7=0.49$

b) both will need repair?

Probability both will need repair = $0.3 \times 0.3=0.09$

c)at least one car will need repair

Neither car will need repair= $0.7 \times 0.7=0.49$

Probability that at least one car will need repair= 1-0.49 = 0.51

6 0

2 years ago