<h2><u>Correct question</u> : 4x - 3 = 2x + 7 </h2><h2><u>Solution</u> : 4x - 3 = 2x + 7 </h2>
Solve for B. R = x(A + B) A) B = (R - A)/x B) B = (A - Rx)/x C) B = (R - Ax)/x
1 answer:
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Answer:
0.85
Step-by-step explanation:
Given two events A and B, the probability that either A or B occurs is given by:
where
the probability that A occurs
is the probability that B occurs
is the probability that both A and B occur at the same time
In this problem, we know the following facts:
is the probability that the car requires an oil change
is the probability that the car requires a brake repair
is the probability that the car requires both an oil change and brake repair
Therefore, the probability that either o (car requiring oil change) or b (car requiring brake repait) occur is:
Answer:
3 + 11a³ - 7a² + 18a - 18
Step-by-step explanation:
<u>When multiplying with two brackets, you need to multiply the three terms, (a²), (4a) and (-6) from the first bracket to all the terms in the second brackets, (3a²), (-a) and (3) individually. I have put each multiplied term in a bracket so it is easier.</u>
(a² + 4a - 6) × (3a² - a + 3) =
(a² × <em>3a²</em>) + {a² × <em>(-a)</em>} + (a² × <em>3</em>) + (4a × <em>3a²</em>) + {4a × <em>(-a)</em>} + (4a × <em>3</em>) + {(-6) × <em>a²</em>) + {(-6) × <em>(-a)</em>} + {(-6) × <em>3</em>}
<u>Now we can evaluate the terms in the brackets. </u>
(a² × 3a²) + {a² × (-a)} + (a² × 3) + (4a × 3a²) + {4a × (-a)} + (4a × 3) + {(-6) × a²) + {(-6) × (-a)} + {(-6) × 3} =
3 + (-a³) + 3a² + 12a³ + (-4a²) + 12a + (-6a²) + 6a + (-18)
<u>We can open the brackets now. One plus and one minus makes a minus. </u>
3 + (-a³) + 3a² + 12a³ + (-4a²) + 12a + (-6a²) + 6a + (-18) =
3 -a³ + 3a² + 12a³ -4a² + 12a -6a² + 6a -18
<u>Evaluate like terms.</u>
3 -a³ + 3a² + 12a³ -4a² + 12a -6a² + 6a -18 = 3
+ 11a³ - 7a² + 18a - 18