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3 years ago

Which expression is a factor of 10x^2+11x+3

Mathematics

1 answer:

Ulleksa [173]3 years ago

5 0

Answer:

The factors are (5x + 3) and (2x + 1)

Step-by-step explanation:

When you need to factor a quadratic, and the coefficient of the x² is not 1, use the slide and divide method.

The general form of a quadratic is ax² + bx + c

Factor: 10x² + 11x + 3

Here a = 10, b = 11, and c = 3

Step 1: Multiply ac, we SLIDE a over to c. Notice the 10 is gone for now..

x² + 11x + 30

Step 2: Factor this (this step will always factor)

x² + 11x + 30 = (x + 5)(x + 6)

So the factors are (x + 5)(x + 6), but we now need to DIVIDE by a, since we multiplied it into c before. We divide the constants in the factors...

(x + 5/10 )(x + 6/10 )

Now reduce the fractions as much as possible...

(x + 1/2 )(x + 3/5)

*If they don't reduce to a whole number, SLIDE the denominator over as a coefficient of x....

(2x + 1)(5x + 3) *2 slide over in front of x, 5 slide over in front of x, the fractions are gone!

These are our factors!

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3 years ago

If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes

Keith_Richards [23]

You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
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X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.

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AlexFokin [52]

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Step-by-step explanation:

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Dafna11 [192]

Answer:

$20a^3b^3$

Step-by-step explanation:

<u>Binomial Series</u>

$(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n$

<u>Factorial</u> is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example: 4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

$\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3$

$\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3$

$\implies \dfrac{6!}{3!3!}a^{3}b^3$

$\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3$

$\implies \left(\dfrac{120}{6}\right)a^{3}b^3$

$\implies 20a^3b^3$

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