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Paraphin [41]
2 years ago
14

Please help with these 2 questions!! (9 and 10) please show with solution needed!!

Mathematics
1 answer:
katrin [286]2 years ago
3 0
9 is m=7
1. 8(m+3)=5(m+9)
2. 8m+24=5m+45
3. get variable on side by its self
3m=21
m=7
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5 – m = 1.36<br><br> m= <br><br> Someone please help me
skelet666 [1.2K]

Answer:

5 - m = 1.36 \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  m = 5 - 1.36 \\ m = 3.64

8 0
3 years ago
A calculator displays a result as 1.3540980 107 kg. the estimated uncertainty in the result is ±2%. how many digits should be i
yan [13]
<span>2 significant digits.
       
Let's see what the range of possible values you can have for 1.3540980 if your uncertainty is +/- 2% 2% of 1.3540980 = 0.02 * 1.3540980 = 0.027082 So the lowest possible value for your result is 1.3540980 - 0.027082 = 1.327016 The largest possible result is 1.3540980 + 0.027082 = 1.38117996 Notice that only the 1st 2 digits of the result match which is reasonable since a 2% error means that your result is only accurate to within 1 part in 50.</span>
3 0
3 years ago
Which side lengths form a right triangle?​
VashaNatasha [74]

Answer:

C

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
You are certain to get 3 jacks when selecting 51 cards from a shuffled deck. Express the indicated degree of likelihood as a pro
Colt1911 [192]

Answer:

0.001145

Step-by-step explanation:

Given that you are certain to get 3 jacks when selecting 51 cards from a shuffled deck.

When we draw 3 cards from 51 cards, we keep one card aside

The card can be either Jack or non Jack

Prob (the left over card to be Jack) = P(A) = \frac{4}{52}

Prob (the left over card to be non jack ) = P(B) = \frac{48}{52}

A and B are mutually exclusive and exhaustive

Let drawing 3 Jacks be the event C

P(C) = P(AC)+P(BC)

P(AC) = P(A)*P(C/A)\\= \frac{4}{52} (3/51C3)\\=1.108(10^{-5}

P(BC) = P(B)*P(C/B)\\

= \frac{28}{52} (4/51C3)\\

=1.034(10^{-4}

Adding we get

P(C)

= (1.108+10.343)(10^{-5} \\=0.001145

3 0
3 years ago
2. Squares are the only regular quadrilaterals.<br> O<br> True<br> O<br> False
Furkat [3]

Answer:

true

Step-by-step explanation:

The only regular (all sides equal and all angles equal) quadrilateral is a square. So all other quadrilaterals are irregular.

8 0
3 years ago
Read 2 more answers
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