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tensa zangetsu [6.8K]
3 years ago
10

What is the ratio of the afternoon to morning temperatures shown in the table?

Mathematics
2 answers:
Monica [59]3 years ago
6 0
I would help you but you need to post the table with the question..
pickupchik [31]3 years ago
5 0

Answer:

32

Step-by-step explanation:

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What is the answer to this problem
In-s [12.5K]
You can do the opposite, as in multiplying.
9.2× 5.3 = 48.76
48.76 = D

5 0
3 years ago
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At 2:00 pm there are 350 freshmen and 200 sophomores at a school carnival. After 2:00 pm 20 freshman leave every hour and 35 sop
dmitriy555 [2]

Answer:

160:340

Step-by-step explanation:

The hours between 2:00pm and 4:00pm is 2 hours. If originally 350 freshman and 200 sophomores are at the carnival, and 20 freshman leave every hour, we can determine how many freshman left in 2 hours:

=20\cdot{2}=40

and if 35 sophomores arrive every half our, we know that for every two hours there is 4 half hours, therefore:

=35\cdot{4}=140

The amount of freshman at 4:00pm:

200-40=160

and the amount of sophomores:

200+140=340

the ratio is 160:340

5 0
3 years ago
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Use Euler’s Formula to find the missing number.<br><br> Faces: _____<br> Edges: 15<br> Vertices: 9
Alla [95]

Answer:

18

Step-by-step explanation:

thanks and follow mark as brainlist

7 0
3 years ago
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Can you answer these please many thanks
givi [52]

Given: (a) 2(x+3)=2x+6 and (b) 4(y-3)=4y-12.

To find: The expanded form of the given expressions.

(c) 4(m+n)=4m+4n       (using a(b+c)=ab+ac)

(d) 3(5-q)=15-3q           (using a(b-c)=ab-ac)

(e) 5(2c+1)=10c+5          (using a(b+c)=ab+ac)

(f) 3(2x-5)=6x-15          (using a(b-c)=ab-ac)

(g) 7(4b-1)=28b-7          (using a(b-c)=ab-ac)

(h) 3(2x+y-5)=3((2x+y)-5)

⇒3(2x+y-5)=3(2x+y)-15         (using a(b-c)=ab-ac)

⇒3(2x+y-5)=6x+3y-15            (using a(b+c)=ab+ac)

(i) 2(6a-4b+3)=2((6a-4b)+3)

⇒2(6a-4b+3)=2(6a-4b)+6         (using a(b+c)=ab+ac)

⇒2(6a-4b+3)=12a-8b+6            (using a(b-c)=ab-ac)

(j)6(m+n+p)=6((m+n)+p)

⇒ 6(m+n+p)=6(m+n)+6p           (using a(b+c)=ab+ac)

⇒ 6(m+n+p)=6m+6n+6p            (using a(b+c)=ab+ac)

(k) y(y+2)=y^2+2y          (using a(b+c)=ab+ac)

(l) g(g-3)=g^2-3g           (using a(b-c)=ab-ac)

(m) n(4-n)=4n-n^2        (using a(b-c)=ab-ac)

(n) a(b+c)=ab+ac           (using a(b+c)=ab+ac)

(o) s(3s-4)=3s^2-4s        (using a(b-c)=ab-ac)

(p) 2x(x+5)=2x^2+10x     (using a(b+c)=ab+ac)

(q) 4y(x-3)=4xy-12y      (using a(b-c)=ab-ac)

(r) 5a(2b-5)=10ab-25a     (using a(b-c)=ab-ac)

(s) 4a(3b+2c)=12ab+8ac    (using a(b+c)=ab+ac)

(t) 5p(4p-5q)=20p^2-25pq    (using a(b-c)=ab-ac)

6 0
3 years ago
Aaron and Belinda went swimming. Aaron swam 575 meters. This is 150 more meters than Belinda swam. Write and solve an addition e
creativ13 [48]

Answer:

Belinda swam 425 meters

Step-by-step explanation:

575=150+x

575-150=x+150-150

425=x

0 0
3 years ago
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