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3 years ago

Which ones have a relationship with functions

Mathematics

2 answers:

egoroff_w [7]3 years ago

8 0

Answer:

b no

c no

d yes

e yes

Step-by-step explanation:

there can not be 2 of the same x's but there can be 2 of the same y's

Anit [1.1K]3 years ago

7 0

Answer:

b. Yes

c. No

d. No

e. Yes

Step-by-step explanation:

hope this helped

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Evaluate the function for the given values to determine if the value is a root. p(−2) = p(2) = The value is a root of p(x).

bija089 [108]

Note: Since you missed to mention the the expression of the function $p(x)$ . After a little research, I was able to find the complete question. So, I am assuming the expression as $p(x)=x^4-9x^2-4x+12$ and will solve the question based on this assumption expression of $p(x)$ , which anyways would solve your query.

Answer:

$p\left(-2\right)=0$

Therefore, $x=-2$ is a root of the polynomial $p(x)=x^4-9x^2-4x+12$

$p\left(2\right)=-16$

Therefore, $x=2$ is not a root of the polynomial $p(x)=x^4-9x^2-4x+12$

Step-by-step explanation:

As we know that for any polynomial let say $p(x)$ , $c$ is the root of the polynomial if $p(c)=0$ .

In order to find which of the given values will be a root of the polynomial, $p(x)=x^4-9x^2-4x+12$ , we must have to evaluate $p(x)$ for each of these values to determine if the output of the function gets zero.

So,

Solving for $p\left(-2\right)$

$p(x)=x^4-9x^2-4x+12$

$p\left(-2\right)=\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12$

$\mathrm{Simplify\:}\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12:\quad 0$

$\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$=\left(-2\right)^4-9\left(-2\right)^2+4\cdot \:2+12$

$\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=a^n,\:\mathrm{if\:}n\mathrm{\:is\:even}$

$=2^4-2^2\cdot \:9+8+12$

$=2^4+20-2^2\cdot \:9$

$=16+20-36$

$=0$

Thus,

$p\left(-2\right)=0$

Therefore, $x=-2$ is a root of the polynomial $p(x)=x^4-9x^2-4x+12$ .

Now, solving for $p\left(2\right)$

$p(x)=x^4-9x^2-4x+12$

$p\left(2\right)=\left(2\right)^4-9\left(2\right)^2-4\left(2\right)+12$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$p\left(2\right)=2^4-9\cdot \:2^2-4\cdot \:2+12$

$p\left(2\right)=2^4-2^2\cdot \:9-8+12$

$p\left(2\right)=2^4+4-2^2\cdot \:9$

$p\left(2\right)=16+4-36$

$p\left(2\right)=-16$

Thus,

$p\left(2\right)=-16$

Therefore, $x=2$ is not a root of the polynomial $p(x)=x^4-9x^2-4x+12$ .

Keywords: polynomial, root

Learn more about polynomial and root from brainly.com/question/8777476

#learnwithBrainly

7 0

3 years ago

Read 2 more answers

In an experiment, some diseased mice were randomly divided into two groups. One group was given an experimental drug, and one wa

damaskus [11]

Answer:

D)Yes, because the difference in the means in the actual experiment was more than two standard deviations from 0.

Step-by-step explanation:

We will test the hypothesis on the difference between means.

We have a sample 1 with mean M1=18.2 (drug group) and a sample 2 with mean M2=15.9 (no-drug group).

Then, the difference between means is:

$M_d=M_1-M_2=18.2-15.9=2.3$

If the standard deviation of the differences of the sample means of the two groups was 1.1 days, the t-statistic can be calculated as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{2.3-(0)}{1.1}=2.09$

The critical value for a two tailed test with confidence of 95% (level of significance of 0.05) is t=z=1.96, assuming a large sample.

This is approximately 2 standards deviation (z=2).

The test statistict=2.09 is bigger than the critical value and lies in the rejection region, so the effect is significant. The null hypothesis would be rejected: the difference between means is significant.

4 0

3 years ago

Diagram below shows the curve of a quadratic function

uranmaximum [27]

Answer:

q = -8, k = 2.

r = -6.

Step-by-step explanation:

f(x) = (x - p)^2 + q

This is the vertex form of a quadratic where the vertex is at the point (p, q).

Now the x intercepts are at -6 and 2 and the curve is symmetrical about the line x = p.

The value of p is the midpoint of -6 and 2 which is (-6+2) / 2 = -2.

So we have:

f(x) = 1/2(x - -2)^2 + q

f(x) = 1/2(x + 2)^2 + q

Now the graph passes through the point (2, 0) , where it intersects the x axis, therefore, substituting x = 2 and f(x) = 0:

0 = 1/2(2 + 2)^2 + q

0 = 1/2*16 + q

0 = 8 + q

q = -8.

Now convert this to standard form to find k:

f(x) = 1/2(x + 2)^2 - 8

f(x) = 1/2(x^2 + 4x + 4) - 8

f(x) = 1/2x^2 + 2x + 2 - 8

f(x) = 1/2x^2 + 2x - 6

So k = 2.

The r is the y coordinate when x = 0.

so r = 1/2(0+2)^2 - 8

= -6.

3 0

3 years ago

A team math contest has $11$ teams of $6$ students each. Before the contest begins, every student high-fives every other student

DedPeter [7]

Question-

A team math contest has $11$ teams of $6$ students each. Before the contest begins, every student high-fives every other student who is not on their team. How many high-fives take place in all?

7 0

3 years ago

165 meters as a percentage of 3 kilometres is

Tems11 [23]

Answer:

165 meters is 5.5% of 3 kilometers.

Solution:

We need to calculate 165 meters is what percent of 3 kilometers.

When comparing or calculating percentage the two quantity should have same units.

So we need to convert 3 kilometers into meters

1 km = 1000 meters

So $1 \times 3 km = 3 \times 1000 meters$

3 km = 3000 meters

Now, need to calculate 165 meters as how much percentage of 3000 meters

So we get

$\frac{165}{3000} \times 100=\frac{11}{2}$ = 5.5%

Hence 165 meters is 5.5% of 3 kilometers.

3 0

3 years ago

Read 2 more answers

Which ones have a relationship with functions ​

Which ones have a relationship with functions