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3 years ago

Determine all values of k that make each of the following a perfect square trinomial. X^2+8x+k

Mathematics

1 answer:

-Dominant- [34]3 years ago

6 0

Answer:

k can either be

−

Step-by-step explanation:

Consider the equation

0=x2+4x+4

. We can solve this by factoring as a perfect square trinomial, so

0=(x+2)2→x=−2 and−2

. Hence, there will be two identical solutions.

The discriminant of the quadratic equation (b2−4ac) can be used to determine the number and the type of solutions. Since a quadratic equations roots are in fact its x intercepts, and a perfect square trinomial will have

2 equal, or 1

distinct solution, the vertex lies on the x axis. We can set the discriminant to 0 and solve:

k2−(4×1×36)=0

k2−144=0

(k+12)(k−12)=0

k=±12

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How do you solve this?

Arada [10]

Make a proportion to get the answer the proportion will be
$\frac{2}{3} = \frac{15}{x}$
to give you the answer of 37.5

6 0

3 years ago

A bottlenose dolphin is 10 feet below sea level. Then it begins to dive at a rate of 9 feet per second. What is the equation of

a_sh-v [17]

Y = -9x - 10
The dolphin is diving 9 feet per second so the rate of change is -9.
It's also 10 feet below sea level which makes the y-intercept -10
Hope this helped!

7 0

3 years ago

Read 2 more answers

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$