Answer:
For f(x) = √(2·x + 2) - √(x + 18), at f(x) = -1 the possible x-values includes;
-0.757, -17.5
Step-by-step explanation:
Given that the function is f(x) = √(2·x + 2) - √(x + 18)
The value of 'x' when f(x) = -1, is given as follows;
-1 = √(2·x + 2) - √(x + 18)
-1² = (√(2·x + 2) - √(x + 18))² = 3·x + 20 - 2·√(2·x + 2)×√(x + 18)
1 = 3·x + 20 - 2·√(2·x + 2)×√(x + 18)
2·√(2·x + 2)×√(x + 18) = 3·x + 20 - 1 = 3·x + 19
2·x² + 38·x + 36 = (3·x + 19)/2
2·x² + 38·x + 36 - (3·x + 19)/2 = 0
4·x² + 73·x + 53 = 0
From which we get;
x = (-73 ± √(73² - 4 × 4 × 53))/(2 × 4)
x ≈ -0.757, and x ≈ -17.5
Answer:
Kaitlin's answer is wrong because she is supposed to by dividing. When trying to get rid of a number from a variable you have to divide the number by itself and whatever you do to one side you have to do to the other side
so you would do 5/5 - cancels out
15/5=3
y=3
Step-by-step explanation:
have a great day :D
Answer:
Hence the probability of the at least 9 of 10 in working condition is 0.3630492
Step-by-step explanation:
Given:
total transistors=100
defective=20
To Find:
P(X≥9)=P(X=9)+P(X=10)
Solution:
There are 20 defective and 80 working transistors.
Probability of at least 9 of 10 should be working out 80 working transistors
is given by,
P(X≥9)=P(X=9)+P(X=10)
<em>{80C9 gives set of working transistor and 20C1 gives 20 defective transistor and 100C10 is combination of shipment of 10 transistors}</em>
P(X≥9)=
<em>(Use the permutation and combination calculator)</em>
P(X≥9)=(231900297200*20/17310309456440)
+(1646492110120/17310309456440)
P(X≥9)=0.267933+0.0951162
P(X≥9)=0.3630492
0
Step-by-step explanation:
for this question is not even possible
easy
Step-by-step explanation:
if you have for example x^2-4
just graf x^2 and that put down all point for 4
