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2 years ago

Find the value of n so that the expression is a perfect square trinomial and then factor the trinomial.

Mathematics

1 answer:

Anna007 [38]2 years ago

7 0

The correct expression is (a) n = 100; (x + 10)^2

<h3>How to determine the value of n?</h3>

The expression is given as:

x^2 + 20x + n

Take the coefficient of x

k = 20

Divide by 2

k/2= 10

Square both sides

(k/2)^2 = 100

The above value represents the value of n

i.e.

n = 100

So, we have:

x^2 + 20x + 100

Expand

x^2 + 10x + 10x + 100

Factorize

x(x + 10) + 10(x + 10)

Factor out x + 10

(x + 10)^2

Hence, the correct expression is (a) n = 100; (x + 10)^2

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If (−4, 11) and (−6, 5) are the endpoints of a diameter of a circle, what is the equation of the circle?

lora16 [44]

Answer:

A. $(x+5)^2+(y-8)^2=100$

Step-by-step explanation:

1. The standard form of the equation of a circle is $(x-h)^{2} + (y-k)^2=r^2$

2. To find the center we require the midpoint of the 2 given points

center = $[\frac{1}{2}(-4-6),\frac{1}{2}(11+5)]$

=(-5,8)

3. The radius is the distance from the centre to either of the 2 given points calculate the radius using the distance formula

d= $\sqrt{(x_{2}+x_{1})^{2} +(y_{2}-y_{1})^{2}$

let $(x_{1},y_{1})= (-5,8) \\$ and $(x_2,y_2)=(-4,11)$

r= $\sqrt{(-4+5)^2+(11-8)^2} =\sqrt{81+9}=10$

--> $(x-(-5)^2+(y-8)^2=10^2$

--> $(x+5)^2+(y-8)^2=100$

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3 years ago

Read 2 more answers

What is the equation of this line in point-slope form AND slope intercept form <br><br> please help

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Point slope form: y-3=-3/5(x+2)
Slope intercept form: y=-3/5x+9/5

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2 years ago

Let Y be a random variable with a density function given by

Neporo4naja [7]

From the given density function we find the distribution function,

$F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y$

(a)

$F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)$

$\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1$

$\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}$

(b)

$F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y$

$\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2$

$\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}$

(c)

$F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})$

$\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3$

$\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}$

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3 years ago

Sylvia ran 3km 290m in the morning then she ran some more in the evening if she ran a total of 10km how far did Sylvia run in th

MAXImum [283]

Answer:

She ran 6.71km in the evening

Step-by-step explanation:

3km 290m = 3.29km

10 - 3.29 = 6.71km

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3 years ago

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Answer:

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2 years ago