3 years ago

-2(6+s) Greater than or equal to -15 - 2s

1 answer:

7 0

Answer:

I'm leaning mostly towards C Because there is a solution, None of them are real numbers and if they were, S would = 5 Especially positive S

~ Zachary

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PLEASE HELP I GIVE THANKS

Levart [38]

A=(-6,-2)
B=(6,2)

$\sqrt{[(6-(-6)]^2+[(2-(-2)]^2} = \sqrt{12^2+4^2} = \sqrt{16} =4 \sqrt{10}$

5 0

3 years ago

9. Choose the correct answer.

denpristay [2]

Answer:

what the h3ll type of question is this ?-

Step-by-step explanation:

your never gonna get one because you dont have any so the probability is 0%.

6 0

3 years ago

Eight years ago the daughter age was thrice the son age

coldgirl [10]

Answer:

what's the question here ?

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3 years ago

In the given figure, O is the centre of a circle, TN is a Tangent, AN = 12 cm and OE = 5 cm. Find the length of NE.

sp2606 [1]

Answer:

NE = 8 cm

Step-by-step explanation:

Radius OA = OE = 5 cm

In right triangle OAN, OA² + AN² = ON²

5² + 12² = ON²

ON² = 25 + 144 = 169

so ON = 13 cm

NE = ON - OE = 13 - 5 = 8 cm

3 0

2 years ago

Which equation could be used to solve this problem? The sum of two consecutive odd integers is 24. Let n represent the first int

USPshnik [31]

If n is the first integer, then n+2 is the second one. The equation for the sum can be written as
n + (n+2) = 24 . . . . . . this equation can be used to solve the problem

2n = 22 . . . . . . . . . . . simplify, subtract 2
n = 11

The integers are 11 and 13.

_____
I like to solve problems like this by looking at the average of the integers. Here, it is 24/2 = 12. The consecutive odd integers whose average is 12 are 11 and 13.

6 0

3 years ago