If you want to find the solutions for this you have to factor it. Since it's a second degree polynomial, you'll have 2 solutions. Factoring this using the quadratic formula, you'll get factors of (5x-8)(3x-4). Solving these for x you get x = 8/5 and x = 4/3.
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Answer:</h3>
C. x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5
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Step-by-step explanation:</h3>
Wrong terms are bolded:
A. x5 + 5x4y + 10x3y2 + 10x3y3 + 5xy4 + y5
B. x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y4
C. x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
D. x5 + 5x4y4 + 10x3y3 + 10x2y2 + 5xy + y5
The sum of exponents should always be 5.
Answer:
y = (6/5) - (7/5) x
Step-by-step explanation:
goal: to rearrange the equation such that it takes the form:
y = f(x) ; where f(x) is an expression in terms of x
Given:
7x+5y=6 subtract 7x from both sides
5y = 6 - 7x (divide both sides by 5)
y = (6-7x) / 5
y = (6/5) - (7/5) x
Answer: (a) 
(b) 
Step-by-step explanation:
(a) P( Bill hitting the target) = 0.7 P( Bill not hitting the target) = 0.3
P( George hitting the target) = 0.4 P(George not hitting the target) = 0.6
Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3
= 0.54
Chances that George hit the target is = 0.4 x 0.3 = 0.12
So given that exactly one shot hit the target, probability that it was George's shot = 
= .
(b) The numerator in the second part would be the same as of (a) part which is 0.12.
The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.
Given that the target is hit,probability that George hit it = 
= =
Answer:
1.) SAS | 2.) HL | 3.) HL | 4.) SSS (Not too sure about this one)
Step-by-step explanation:
