Answer:
Down Here ↓↓↓↓↓↓↓
Step-by-step explanation:
1/8 = 1 ÷ 8
It might be weird with the formatting but here:
0. 1 2 5
____________
8 | 1. 0 0 0
− 0
____________
1 0
− 8
____________
2 0
− 1 6
___________
4 0
− 4 0
___________
0
-Chetan K
Answer:
a= v+w-k
Step-by-step explanation:
Get 'a' by itself to make it a solvable equation
so, subtract k from both sides so it can cancel itself on the 'a' side
Isosceles and a right triangle
Problem 1
<h3>Answer: choice B) They are supplementary angles</h3>
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Explanation:
The two angles combine to form a straight angle (which is 180 degrees). Another term for these angles is "linear pair" since they form a straight line.
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Problem 2
<h3>Answer: choice B) x = 7</h3>
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Explanation:
The diagonal is never used in this problem. For any rhombus, the four exterior sides are always the same length.
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Problem 3
<h3>Answer: Choice D) 62 degrees</h3>
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Explanation:
We use the angle addition postulate. This allows us to combine angles BAC and CAD to form angle BAD.
(angle BAC)+(angle CAD) = angle BAD
30 + x = 92
x = 92-30
x = 62
angle CAD = 62
The weight of the new student is 27 kg.
Average weight
= total weight ÷total number of students
<h3>
1) Define variables</h3>
Let the total weight of the 35 students be y kg and the weight of the new student be x kg.
<h3>2) Find the total weight of the 35 students</h3>
<u>
</u>
y= 35(45)
y= 1575 kg
<h3>3) Write an expression for average weight of students after the addition of the new student</h3>
New total number of students
= 35 +1
= 36
Total weight
= total weight of 35 students +weight of new students
= y +x
![44.5 = \frac{y + x}{36}](https://tex.z-dn.net/?f=44.5%20%20%3D%20%20%5Cfrac%7By%20%2B%20x%7D%7B36%7D%20)
<h3>4) Substitute the value of y</h3>
![44.5 = \frac{1575 + x}{36}](https://tex.z-dn.net/?f=44.5%20%3D%20%20%5Cfrac%7B1575%20%2B%20x%7D%7B36%7D%20)
<h3>5) Solve for x</h3>
36(44.5)= 1575 +x
1602= x +1575
<em>Subtract 1575 from both sides:</em>
x= 1602 -1575
x= 27
Thus, the weight of the new student is 27 kg.