(5x)^2 + x^2 = c^2
25x^2 + x^2 =c^2
sqrt 26 x ^ 2 = sqrt c^2
x sqrt 26= c
so answer is 6x + x sqrt 26
Answer:
y=3x
Step-by-step explanation:
y=mx+b, m is slope, b is y-intercept
Answer:
a) ⅓ units²
b) 4/15 pi units³
c) 2/3 pi units³
Step-by-step explanation:
4y = x²
2y = x
4y = (2y)²
4y = 4y²
4y² - 4y = 0
y(y-1) = 0
y = 0, 1
x = 0, 2
Area
Integrate: x²/4 - x/2
From 0 to 2
(x³/12 - x²/4)
(8/12 - 4/4) - 0
= -⅓
Area = ⅓
Volume:
Squares and then integrate
Integrate: [x²/4]² - [x/2]²
Integrate: x⁴/16 - x²/4
x⁵/80 - x³/12
Limits 0 to 2
(2⁵/80 - 2³/12) - 0
-4/15
Volume = 4/15 pi
About the x-axis
x² = 4y
x² = 4y²
Integrate the difference
Integrate: 4y² - 4y
4y³/3 - 2y²
Limits 0 to 1
(4/3 - 2) - 0
-2/3
Volume = ⅔ pi
Complete question :
Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.
What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.
Answer:
0.56 ; 0.60
Step-by-step explanation:
From The attached Venn diagram :
C = attend college ; J = has a job
P(C) = (35+45)/300 = 80/300 = 8/30
P(J) = (30+45)/300 = 75/300 = 0.25
P(C n J) = 45 /300 = 0.15
1.)
P(J | C) = P(C n J) / P(C)
P(J | C) = 0.15 / (8/30)
P(J | C) = 0.5625 = 0.56
2.)
P(C | J) = P(C n J) / P(J)
P(C | J) = 0.15 / (0.25)
P(C | J) = 0.6 = 0.60


<- Distributive Property

<- Combine Like Terms
If you're trying to solve for 0:


<- Subtracted 18 from both sides

<- Divided both sides by 60 and then simplified.

<- Fraction Form

<- Decimal Form
Give Brainliest for simple answer plz :P