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QveST [7]
3 years ago
15

Carmen went rock climbing at the climbing gym for 1 1/2 hours on Thursday. On Saturday, she went to Castle Rock State Park and c

limbed for 1 1/2 times as long. How much time did she spend rock climbing in all?
Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Answer:

3 3/4 hours

Step-by-step explanation:

1 1/2 + 1 1/2(1 1/2) = 3 3/4 hours

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Rewrite 2 + (x + 4) as an equivalent expression that doesn't use parentheses
Vladimir79 [104]
2+x+4 just drop the parenthasesi think
5 0
4 years ago
F (x) when x = 1 for the function f (x) = −x2 − 3.
zepelin [54]
The answer is - x 2-3
put x=1
(-1)2-3
-2 -3
=-5
4 0
4 years ago
HELP
charle [14.2K]

Answer:

2

Step-by-step explanation:

So let's call the time she needs to catch up "x", and so Terry's rate would be x-6 since he left home 6 hours earlier. Then we multiply the rates to get 20x - 120 = 80x. The answer is -2, but we can just say 2.

Checking our answer, that means if it took Sally 2 hours, it took Terry 8 hours for he left home 6 hours earlier. So 8 times 20 = 160km traveled. 2 times 80 = 160km traveled. Same distance.

3 0
3 years ago
Solve the inequality.
frutty [35]

Answer:

\boxed{\boxed{\sf x>-7}}

Step-by-step explanation:

\sf -5x-7

Add 7 to both sides:

  • \sf -5x-7+7
  • \sf -5x

Multiply both sides by -1:

  • \sf \left(-5x\right)\left(-1\right)>35\left(-1\right)
  • \sf 5x>-35

Divide both sides by 5:

  • \sf \cfrac{5x}{5}>\cfrac{-35}{5}
  • \sf x>-7

_________________________________

3 0
3 years ago
The perpendicular bisectors of sides AC and BC of △ABC intersect side AB
tiny-mole [99]

The measure of ∠ACB will be 110°

<u><em>Explanation</em></u>

According to the diagram below, DE and DF are the perpendicular bisectors of AC and BC respectively and they intersect side AB at points P and Q respectively.

So, AE=CE and CF= BF

Now, <u>according to the SAS postulate</u>, ΔAPE and ΔCPE are congruent each other. Also, ΔCFQ and ΔBFQ are congruent to each other.

That means, ∠PCE = ∠PAE  and ∠FCQ = ∠FBQ

As ∠CPQ = 78° , so  ∠PCE + ∠PAE = 78°  or,  ∠PCE = \frac{78}{2}= 39°                                    and as ∠CQP = 62° , so ∠FCQ + ∠FBQ = 62° or, ∠FCQ = \frac{62}{2}=31°

Now, in triangle CPQ,  ∠PCQ = 180°-(78° + 62°) = 180° - 140° = 40°

Thus, ∠ACB = ∠PCE + ∠PCQ + ∠FCQ = 39° + 40° + 31° = 110°



7 0
3 years ago
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