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Flura [38]
3 years ago
13

Can someone help me with N please

Mathematics
1 answer:
oksian1 [2.3K]3 years ago
7 0
Let's assume x Mo's number.

11+x/3=-1
==>x/3=-1-11
==>x/3=-12
==>x=-12*3
==>x=-36

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Linear functions have no exponents higher than 1, and a graph that looks like a straight line. non-linear functions have at least one exponent higher than 1, and a graph that isn't a straight line
8 0
2 years ago
I'LL GIVE BRAINLIEST IF YOU EXPLAIN THE ANSWER:
USPshnik [31]

Answer:

C) 16 cm^2

Step-by-step explanation:

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2 years ago
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What is the difference?
matrenka [14]

Answer:

The Difference in mathematical terms is the result of subtracting one number from another.

Step-by-step explanation:

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8 0
2 years ago
Tony separates a number of paintbrushes, x, into 3 equally sized groups. He claims that this is the same as making a group 1/3 t
DochEvi [55]

Answer:

Size of each group = \frac{x}{3}

Step-by-step explanation:

Total number of paint brushes = x

Since each the three groups are equally sized, to get the number of paintbrushes in a group, we will have to divide the total number of paint brushes by 3

∴ We have, number of paint brushes in a group = \frac{Total number of paintbrushes}{3}\\

= \frac{x}{3}

this is also the same thing as saying \frac{1}{3}×\frac{x}{1}

This shows that the size of the new group is the same as \frac{1}{3} the size of the original group

4 0
3 years ago
​Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.
Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

x^{3}+7 x^{2}+12 x=0

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

x^{3}+7 x^{2}+12 x=0

x\left(x^{2}+7 x+12\right)=0   ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

x^{2}+7 x+12=0

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}

<em><u>Therefore, the two roots are:</u></em>

\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}

And,

\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0
3 years ago
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