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Softa [21]
3 years ago
10

in a class of 30 students the highest score in physics test was 98 and the lowest was 35 what was the range?​

Mathematics
1 answer:
In-s [12.5K]3 years ago
5 0

Answer:

63

Step-by-step explanation:

basta ayon sagot ko obob ko eh

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Need help with this someone send the answer please i don't understand why there are so many boxes.
Inessa05 [86]

Answer: See the attached image below. I've filled in the table with the proper values.

Row 2: AFB = 36, BFC = 54, EFA = 144

Row 3: AFB = 40, BCF = 50, DFE = 40

=======================================================

Explanation:

angles AFB and BFC add to 90 degrees since they form a right angle.

In row 1, we see that 48+42 = 90

Angles AFB and DFE are congruent since they are vertical angles. For row 1, we have both of these angles equal to 48 degrees.

Also for row 1, we have angles BFC and CFD add to 42+90 = 132, which is exactly the measure of angle BFD. This angle is congruent to angle EFA because they are vertical angles. So that's how angle EFA is 132.

--------------------------

Now onto row 2

We'll use the ideas of row 1 to work our way backwards to fill in the missing items.

We're given DFE is 36 degrees. This must mean that AFB is also 36 degrees. These angles are vertical angles.

Then we'll use the idea that (angleAFB)+(angleBFC) = 90 to find that

angle BFC = 90 - (angle AFB) = 90 - 36 = 54

Lastly for this row, we can say,

angle BFD = (angle BFC) + (angle CFD) = 54 + 90 = 144

angle EFA = angle BFD = 144

To summarize the missing items of row 2, we have...

  • angle AFB = 36
  • angle BFC = 54
  • angle EFA = 144

---------------------------

Now onto row 3. We use the same tricks (more or less) that were used on row 2.

If angle EFA is 140, then so is angle BFD. These are vertical angles.

(angle BFC) + (angle CFD) = angle BFD

(angle BFC) + 90 = 140

angle BFC = 140-90

angle BFC = 50

This in turn lets us say,

(angle AFB) + (angle BFC) = 90

angle AFB = 90 - (angle BFC)

angle AFB = 90 - 50

angle AFB = 40

angle DFE = angle AFB = 40

To summarize the missing pieces we found:

  • angle AFB = 40
  • angle BFC = 50
  • angle DFE = 40

All of the values are filled in the table shown below.

7 0
3 years ago
What is The answer for number one?
maks197457 [2]
I think is 22
sorry if i'm wrong 
4 0
3 years ago
What command will bring Kyle back to his original size?
sashaice [31]
Back to your original size
3 0
4 years ago
What is 20% of 62.50
MAXImum [283]
It would be:

12.5 

Hope this helped, If I am wrong sorry, I have not done this in a while
7 0
3 years ago
Please explain how to do the problem not just give the answer will mark brainiest
goldfiish [28.3K]
1 Convert
5
2
3
5
​3
​
​2
​​ to improper fraction. Use this rule:
a
b
c
=
a
c
+
b
c
a
​c
​
​b
​​ =
​c
​
​ac+b
​​
5
×
3
+
2
3
−
3
3
4
=
1
+
2
3
+
3
4
​3
​
​5×3+2
​​ −3
​4
​
​3
​​ =1+
​3
​
​2
​​ +
​4
​
​3
​​

2 Simplify
5
×
3
5×3 to
1
5
15
1
5
+
2
3
−
3
3
4
=
1
+
2
3
+
3
4
​3
​
​15+2
​​ −3
​4
​
​3
​​ =1+
​3
​
​2
​​ +
​4
​
​3
​​

3 Simplify
1
5
+
2
15+2 to
1
7
17
1
7
3
−
3
3
4
=
1
+
2
3
+
3
4
​3
​
​17
​​ −3
​4
​
​3
​​ =1+
​3
​
​2
​​ +
​4
​
​3
​​

4 Convert
3
3
4
3
​4
​
​3
​​ to improper fraction. Use this rule:
a
b
c
=
a
c
+
b
c
a
​c
​
​b
​​ =
​c
​
​ac+b
​​
1
7
3
−
3
×
4
+
3
4
=
1
+
2
3
+
3
4
​3
​
​17
​​ −
​4
​
​3×4+3
​​ =1+
​3
​
​2
​​ +
​4
​
​3
​​

5 Simplify
3
×
4
3×4 to
1
2
12
1
7
3
−
1
2
+
3
4
=
1
+
2
3
+
3
4
​3
​
​17
​​ −
​4
​
​12+3
​​ =1+
​3
​
​2
​​ +
​4
​
​3
​​

6 Simplify
1
2
+
3
12+3 to
1
5
15
1
7
3
−
1
5
4
=
1
+
2
3
+
3
4
​3
​
​17
​​ −
​4
​
​15
​​ =1+
​3
​
​2
​​ +
​4
​
​3
​​

7 Find the Least Common Denominator (LCD) of
1
7
3
,
1
5
4
​3
​
​17
​​ ,
​4
​
​15
​​
Method 1: By Listing Multiples
List out all multiples of each denominator, and find the first common one.
Multiples of 3 : 3, 6, 9, 12, ...
Multiples of 4 : 4, 8, 12, ...
Therefore, the LCD is
1
2
12.
Method 2: By Prime Factors
List all prime factors of each denominator, and find the union of these primes.
Prime Factors of 3 : 3
Prime Factors of 4 : 2, 2
Therefore, the LCD is
2
×
2
×
3
=
1
2
2×2×3=12.
3 0
3 years ago
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