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Sav [38]
3 years ago
10

-3 (2m+5)+8=5 (-2m+1)

Mathematics
2 answers:
asambeis [7]3 years ago
7 0
-4,5 please my answer
Zanzabum3 years ago
3 0
I just worked out the problem twice and got -4.5 both times.
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nancy will arrive at the hotel on July 8, and will stay three nights. What date will Nancy check out of the hotel?
Inessa [10]

Answer:

july 11

Step-by-step explanation:

6 0
3 years ago
Solve x+2/x-4&lt;0<br><br> A)–4 &lt; x &lt; –2<br> B)–2 &lt; x &lt; 4<br> C)–2 &lt; x &lt; –4
Gelneren [198K]

Domain:\ x\neq-2\\\\\dfrac{x+2}{x-4}

Other method:

\dfrac{x+2}{x-4}

zeros of numerator and denominator are x = -2 and x = 4.

Look at the second picture.

for x < -2 → x + 2 < 0 and x - 4 < 0

therefore \dfrac{x + 2}{x - 4}=\dfrac{(-)}{(-)}>0

for -2 < x < 4 → x + 2 > 0 and x - 4 < 0

therefore \dfrac{x+2}{x-4}=\dfrac{(+)}{(-)} < 0

for x > 4 → x + 2 > 0 and x - 4 > 0

therefore \dfrac{x+2}{x-4}=\dfrac{(+)}{(+)} > 0

<h3>Answer: B) -2 < x < 4</h3>

4 0
3 years ago
Find the length of “b”, to the<br> nearest tenth, using the<br> Pythagorean Theorem.
Tresset [83]

Answer: b= 11.2

10^2 + b^2 = 15^2

100 + b^2 =  225

b^2 = 125

b= \sqrt{125}

b= 11.2

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
11. Given: TQ bisects ZRTP, QS || PT<br> RT = 30. Find QP, RS and QS
yaroslaw [1]

QP=24 cm

RS=11.25 cm

QS=18.75 cm

<u>Explanation</u>:

Given that TQ bisects <RTP

therefore(1)

consider ΔRQS and ΔRPT

QS||PT,RP and RT are transversals

(alternate angles)(2)

comparing (1) and (2)

and triangle SQT is isocelus

Therefore SQ=ST(sides opposite to equal angles in an isocelus triangle)

Therefore <RQS=<RPT(corresponding angles)

<RSQ=<RTP(corresponding angles)

therefore by AA criterion for similarity

ΔRQS~ΔRPT

According to the property of similar triangles

RQ/RP=RS/RT=QS/PT9/RP=X/30=QS/50\\9/RP=X/30=(30-X)/50\\X/30=(30-X)/50\\50X=30(30-X)\\50X=900-30X\\50X+30X=900\\80X=900\\X=900/80=11.25\\RS=11.25 cm\\QS=30-X=30-11.25\\QS=18.75 cm\\9/RP=X/30\\9/RP=11.25/30\\9*30/11.25=RP\\RP=24 cm

3 0
3 years ago
Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen
sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

5 0
3 years ago
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