Answer:
101
Step-by-step explanation:
Given the expression:
2016 · 199 = 2016·? +2016 · 98
We could rewrite the expression such that the missing value equals x
2016 * 199 = 2016 * x + 2016 * 98
Using BODMAS:
401184 = 2016x + 197568
401184 - 197568 = 2016x
203616 = 2016x
Divide both sides by 2016
203616 / 2016 = 2016x / 2016
101 = x
Hence, missing value is 101
9 inches * 1 foot / 12 inches
= 3/4 feet deep
Area of the sidewalk:
= the sides of the pool * the width of the sidewalk + the corners of the sidewalk (width * width)
= 2 sides * 6 (feet / side) * 2 feet + 2 sides * 14 (feet / side) * 2 feet + 4 corners * (2 feet * 2 feet / corner)
= 12 feet * 2 feet + 28 feet * 2 feet + 4 * 2 feet * 2 feet
= 24 feet^2 + 56 feet^2 + 16 feet^2
= 96 feet^2
Volume:
= 96 feet^2 * 3/4 feet
= (96 * 3 / 4) feet^3
= (19 * 3) feet^3
= 57 feet^3
Answer:
Multiply vector c by the scalar -1/2.
Step-by-step explanation:
Look at vector c.
It has an x component of 4 and a y component of 4.
You can write vector c as a sum of its components using unit vectors in the x direction (i) and in the y direction (j).
c = 4i + 4j
Now look at vector d, and write it also as a sum of its x and y components.
d = -2i - 2j
Now ask yourself, what operation do I do to 4 to end up with -2?
One answer is to multiply 4 by -1/2.
d = (-1/2)c = (-1/2)(4i) + (-1/2)(4j) = -2i - 2j
That worked. By multiplying vector c by the scalar -1/2, you end up with vector d.
Answer:
Step-by-step explanation:
1) rewrite 60 as it's prime factors.
2) group the smae prime factors into pairs.
3) Rewrite each pair in exponent form.
4) use this role:
5) Simplify.
So, therefor, the answer is option 1.
The average rate of change of the function in the given interval -3 ≤ x ≤-1 is 10.
<h3>What is the average rate of change?</h3>
The average Rate of Change of the function f(x) can be calculated as;
Interval -3 ≤ x ≤-1
a = -3, b = -1
f(a) = f(-3) = -20
f(b) = f(-1) = 0
Step 2: Find Average
Therefore, the average rate of change of the function in the given interval -3 ≤ x ≤-1 is 10.
Learn more about average rate;
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