To find perimeter you add up all the sides so the answer is 210
Answer:
0.64 seconds
Step-by-step explanation:
In the equation provided:
h = −16t2 + 4t + 4
h is the height of the ball and t is time. Since we want to find the time when the ball touches the floor, then height is 0. This leaves us with the equation
-16
+ 4t + 4 = 0
This is a quadratic equation can be solved with the following formula:
where a=-16
b=4
c=4
Solving for t we will find two different results:
Since time can't be negative, we discard t2 and choose t1.
Since it is required to answer in the nearest hundredth, we round the result to t=0.64 seconds.
The Set Up:
x² = (Side1)² + (Side2)² - 2[(Side1)(Side2)]
Solution:
cos(Toby's Angle) • x² = 55² + 65² - 2[(55)(65)] cos(110°)
x² = 3025 + 4225 -7150[cos(110°)]
x² = 7250 - 2445.44x =
√4804.56x = 69.31m
The distance, x, between two landmarks is 69.31m.
Note: The answer choices given are incorrect.
The answer is 41 inches. Since <span>△HKS∼△DLY, so the measurements of the angles are just the same. So if the sides are 15 x 20 x 32. Then to get the sides of the other triangle is simply adding what is the measurement of the sides from the first triangle. so it is 24 x 29 x 41. So the answer of DY is equal to 41 inches.</span>
Two fractions equivalent to each: Just divide or multiply both top AND bottom by the same number.<span>
5/6: 10/12 OR 15/18
15/30: 5/10 OR 1/2
45/60: 8/12 OR 4/6
Rewrite each pair or fractions with common denominator: Find the difference between the two bottom numbers, and multiply top and bottom number.
5/8 and 3/4: 4X2=8, 3X2=6. So, 5/8 and 6/8.
2/5 and 1/2: 2/5 and 2.5/5
9/9 and 5/7: 9/9 and ~5.7/9
Rewrite each in simple form: Find greatest common factor and divide.
9/54: 1/6
20/40: 1/2
100/110: 10/11
Are these fractions equivalent?
No. 5/1 and 5/5 are, because they are both 5 wholes. 1/5 is not because it is a fifth of a whole.
In what situation can you use multiplication to find equivalent fractions?
I'm sorry but I do not understand this question.
</span>Source(s):<span>I hope I helped, seeing as I have graduated with a math degree.</span>
