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igor_vitrenko [27]

2 years ago

13

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each situation to its corresponding

expression.

1 answer:

dlinn [17]2 years ago

5 0

Answer:

I mean how can I answer this???????????????????????

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Solve the following absolute value equation.<br><br> −8 − 2 |5X – 17| = −14

Levart [38]

If 5x - 17 ≥ 0, then |5x - 17| = 5x - 17 and the equation reduces to

-8 - 2 (5x - 17) = -14

-8 - 10x + 34 = -14

-10x = -40

x = 4

If 5x - 17 < 0, then |5x - 17| = -(5x - 17) = -5x + 17 and we have

-8 - 2 (-5x + 17) = -14

-8 + 10x - 34 = -14

10x = 28

x = 2.8

4 0

1 year ago

laila [671]

F(x)=x^3-9x
and
g(x)=x^2-2x-3

so you just need to divide f(x) by g(x)

Therefore:

f(x)/g(x) = (x^3-9x) / (x^2-2x-3)

and of course you need to factor these two function to see if some factor would cancel another

x^3-9x = x(x^2-9)=x(x-3)(x+3)
and
x^2-2x-3 = (x-3)(x+1)

8 0

2 years ago

A tennis team has won 15 out of 20 matches they played this season. How many additional matches must the team win in a row to ra

LuckyWell [14K]

You need to win 30 matches in a row to raise its winning percentage to 90%

7 0

3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
\\\\\\
\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
\\\\\\
-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
\\\\\\
\textit{the initial \underline{position} was 400ft away at 0secs}
\\\\\\
-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

$\bf s(t)=-16t^2+400\implies 0=-16t^2+400\implies \cfrac{-400}{-16}=t^2
\\\\\\
25=t^2\implies \boxed{5=t}$

part B) check the picture below

5 0

2 years ago

Identify the interquartile range for the set of values below. 37, 34 , 29 , 33, 35 , 30 , 34 A) 4 B) 5 C) 3 D) 8

dusya [7]

Answer:25th Percentile: 5

50th Percentile: 30

75th Percentile: 34

Interquartile Range: 29

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers