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3 years ago

7

What is a 8096 rounded to 1 significant figure

2 answers:

guajiro [1.7K]3 years ago

5 0

Answer:

Y Me Angry U Hungry Me And Me Angry Way Back together Me Y He Angry Me He Angry Me Angry:(

d1i1m1o1n [39]3 years ago

5 0

Answer is 8000
8 is the first non-zero digit leaving it the one used

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In a 30-60-90 triangle, what is the length of the hypotenuse when the shorter leg is 5 cm?

zheka24 [161]

Shorter Leg=x

Hypotenuse=2x

Shorter Leg=5cm

Hypotenuse= 10cm

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3 years ago

Please help thank you

OleMash [197]

Answer:

your answer for K should be 12

8 0

4 years ago

Read 2 more answers

Helpppp please any answers

hoa [83]

Step-by-step explanation:

2 + 3 × 6 + 4 × 3

Here, Parantheses should be put,

<u>(2 + 3) </u>× 6 + 4 × 3. (P.E.D.M.A.S) ( First Parantheses)

= 5 × 6 + 4 × 3 (Then multiplication)

= 30 + 12. (At last addition)

= 42

7 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]

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

Answer:

As

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

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

As

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

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

Step-by-step explanation:

As we know that for any polynomial let say<em> </em> $p(x)$ <em>, </em> $c$ 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, $p(x)=x^4-9x^2-4x+12$ <em>, </em>we must have to evaluate <em> </em> $p(x)$ <em> </em>for each of these values to determine if the output of the function gets zero.

So,

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

<em> </em> $p(x)=x^4-9x^2-4x+12$

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

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

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

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

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

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

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

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

$=16+20-36$

$=0$

Thus,

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

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

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

<em> </em> $p(x)=x^4-9x^2-4x+12$

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

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

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

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

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

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

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

Thus,

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

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

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

Find the missing exponent.

777dan777 [17]

Answer: what missing exponent?

Step-by-step explanation:

7 0

3 years ago