What is the area of a rectangular garden that is 60 feet by 20 feet

A number is divided by 3. That quotient is increased by 12. The result is 46. Find the number.

anzhelika [568]

Let the number be x.

x/3 + 12 = 46

Subtract 12 from both sides and you will get:

x/3 = 34

Multiply both sides by 3:

x = 102

3 0

3 years ago

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:

As

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

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

As

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

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

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7 0

3 years ago

Read 2 more answers

Answer two questions about Systems A and B:System A x+3y=−9 2x+y=4 System B 3x+4y=−9 2x+y=4

Jet001 [13]

Answer:

A=(21/5,-22/5) B=(5,-6)

Step-by-step explanation:

6 0

2 years ago

Write an equation of the line with the given slope and y-intercept.

11Alexandr11 [23.1K]

Y=-1/3X-1 should be correct

4 0

3 years ago

Study the Proof: Give the reasons for the missing reason. List the number and then state the reason.

joja [24]

1) Given
2) Adding segments AB and BC together gives AC (Addition postulate, not too sure about the terminology)
3) Segment Addition postulate
4) Segment Addition postulate(Knowing that AC=BD, the distance between BC and CD add up to BD, shown in #3. AB and BC add together to give AC)
5) Subtraction property of Equality

Hope I helped :)

6 0

3 years ago