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3 years ago

How do I factor this polynomial? q^2-121

Mathematics

1 answer:

Tresset [83]3 years ago

5 0

$q^2-121 =q^2-11^2=(q-11)(q+11)\\ \\ \\a^2-b^2 =(a-b)(a+b)$

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KIM [24]

Answer:

x=-1 is the answers for the question

Step-by-step explanation:

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2 years ago

What is M ∠KNL ? Circles

alex41 [277]

Answer:

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3 years ago

At the U.S open tennis championship a statistician keeps track of every serve that a player hits during the tournament. The mean

GarryVolchara [31]

Answer:

Step-by-step explanation:

Both 115 and 145 mph are above the mean. Draw a normal curve and mark these speeds. 115 mph is 1 standard deviation above the mean; 130 would be 2 standard deviations above the mean; and 145 would be 3 s. d. above it.

We need to find the area under the standard normal curve between 115 and 145. This is equivalent to the area under the standard normal curve between z = 1 and z = 3.

I used my TI-83 Plus calculator's DISTR function "normalcdf(" to calculate this area: normalcdf(1, 3) = 0.1573.

The area between z = 1 and z = 3 is 0.1573. In other words, the percentage of serves that were between 115 and 145 mph was 15.73%.

8 0

3 years ago

A line passes through the given points. Write an equation for the line in point-slope form. Then rewrite the equation in slope-i

Andre45 [30]

Answer:

The answer to your question is a) y - 2 = -5(x - 3)

b) y = -5x + 17

Step-by-step explanation:

Data

A (1, 12)

B (3, 2)

Process

1.- Find the slope

Formula

$m = \frac{y2 - y1}{x2 - x1}$

Substitution

$m = \frac{2 - 12}{3 - 1} = \frac{-10}{2} = - 5$

2.- Find the equation in point slope form

Equation

y - y1 = m(x - x1)

Substitution and equation

y - 2 = -5(x - 3)

3.- Find the slope-intercept form

Expand the point slope form

y - 2 = -5x + 15

Simplify

y = -5x + 15 + 2

Equation

y = -5x + 17

6 0

3 years ago

Read 2 more answers

Solve 2x^2 + x - 4 = 0 <br> X2 +

damaskus [11]

Answer:

$\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }$

Step-by-step explanation:

Hello, please find below my work.

$2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0$

$\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0

3 years ago