V = PI x r^2 x H
1348.79 = 3.14 x 5.5^2 x H
1348.79 = 94.985 x h
h = 1348.79 / 94.985
h = 14.2 cm ( round answer if needed)
Answer: True
Step-by-step explanation:
If x2 = 49, then x = 7. True; x = 7 if and only if x^2 = 49.
I hope this solves your problem! It was a little tricky but I tried.
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Answer:
y(s) = 
we will compare the denominator to the form 

comparing coefficients of terms in s
1
s: -2a = -10
a = -2/-10
a = 1/5
constant: 

hence the first answers are:
a = 1/5 = 0.2
β = 5.09
Given that y(s) = 
we insert the values of a and β
= 
to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A
5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)
-52 = B(26)
B = -52/26 = -2
to get A lets substitute s=0.4
5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)
-51 = 0.2A - 52.08
0.2A = -51 + 52.08
A = -1.08/0.2 = 5.4
<em>the constants are</em>
<em>a = 0.2</em>
<em>β = 5.09</em>
<em>A = 5.4</em>
<em>B = -2</em>
<em></em>
Step-by-step explanation:
- since the denominator has a complex root we compare with the standard form

- Expand and compare coefficients to obtain the values of a and <em>β </em>as shown above
- substitute the values gotten into the function
- Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
- after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above
Thanks...
Answer: d=20/3
Step By Step:
Step 1, Simplify:
12-3/4(d+16)=-5
(-3/4)d+(12-12)=-5
(-3/4)d=-5
Step 2, Multiply each side by 4/(-3)
(4/-3)*(-3/4)d=(4/-3)*-5
d=20/3
Answer:
6.31 mi
Step-by-step explanation:
The diagram below explains the solution better.
From the diagram,
C = starting point of the race.
A = end of the first part of the race.
B = end of the race.
Using Cosine rule, we can find the straight-line distance between the starting point and the end of the race.
Cosine rule states that:
![a^2 = b^2 + c^2 - 2bc[cos(A)]](https://tex.z-dn.net/?f=a%5E2%20%3D%20b%5E2%20%2B%20c%5E2%20-%202bc%5Bcos%28A%29%5D)
where A = angle A = <A
Given that
b = 5.2 miles
c = 2.0 miles
<A = 115° (from the diagram)
Hence,
![a^2 = 5.2^2 + 2.0^2 - 2*5.2*2.0[cos(115)]\\\\a^2 = 27.04 + 4 - 20.8[cos(115)]\\\\a^2 = 31.04 + 8.79\\\\a^2 = 39.83\\\\a = \sqrt{39.83}\\ \\a = 6.31 mi](https://tex.z-dn.net/?f=a%5E2%20%3D%205.2%5E2%20%2B%202.0%5E2%20-%202%2A5.2%2A2.0%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2027.04%20%2B%204%20-%2020.8%5Bcos%28115%29%5D%5C%5C%5C%5Ca%5E2%20%3D%2031.04%20%2B%208.79%5C%5C%5C%5Ca%5E2%20%3D%2039.83%5C%5C%5C%5Ca%20%3D%20%5Csqrt%7B39.83%7D%5C%5C%20%5C%5Ca%20%3D%206.31%20mi)
The straight-line distance between the starting point and the end of the race is 6.31 mi