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kramer
3 years ago
6

Plz i need a quick help doing this question

Mathematics
1 answer:
Anni [7]3 years ago
6 0
-13 degrees celsius
You might be interested in
PLEASE HELP! Find an equation of the line whose intercepts are twice those of the graph of 5y+2x=10.
valina [46]

5y+2x=10y=−25x+2246810−2−4−6−8−10246810−2−4−6−8−10Let's solve for x.5y+2x=10Step 1: Add -5y to both sides.2x+5y+−5y=10+−5y2x=−5y+10Step 2: Divide both sides by 2.2x2=−5y+102x=−52y+5Answer:x=−52y+5

6 0
3 years ago
Read 2 more answers
A: 28, 23,30, 25, 27
Serggg [28]

Answer:

8

Step-by-step explanation:

mean of A=28+23+30+25+27/5 = 26.6

average of B=22+19+15+17+20/5 = 18.6

now,

difference= 26.6-18.6

=8,,

6 0
3 years ago
10×10×25= show your work please
zalisa [80]
10 ×10 = 100
100 ×25 = 2500
so your answer is 2500
3 0
3 years ago
Read 2 more answers
4 bells toll together at 9 am. They toll after 7, 8. 11 and 12 seconds respectively. How many times will they toll together agai
Svetach [21]

Answer:

5 times

Step-by-step explanation:

What we need to do here is to find the lowest common multiple of 7,8,11 and 12

This is;

1848

Now, let’s calculate the number of seconds in 3 hours

In 1 hour, we have 3600 seconds

In 3 hours, this will be;

3600 * 3 = 10,800 seconds

So, to find the number of times in which all 4 will toll together, we need to know the number of 1848 seconds in 10,800 seconds

Mathematically, that will be;

10,800/1848 = 5.844

Our answer will be 5 times however

This is because the sixth time they would have sounded together would be more than the 3 hours time frame

8 0
3 years ago
Can u guys answer my question 13 and 14 pls
Alika [10]

Answer:

√2=1.414

then :√8 +2√32 +3√128+4√50

√8=√2³ =2√2

2√32=√2^5 = 4*2√2 = 8√2

3√128 = 3√2^6*2=8*3√2 =24√2

4√50 =4√5²*2= 20√2

add results : 2√2+8√2 +24√2+20√2=54√2

<h2>54√2=54×1.414=76.356 ( it is not in the options)</h2>

x=7-4√3

√x+ 1/√x

√(7-4√3) +1/√(7-4√3) =

(8-4√3)/√(7-4√3)

(8-6.93)/√(7-6.93) = 4 ( after rounded to the nearest whole number)

4 is your answer

3 0
3 years ago
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