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2 years ago

Find the perimeter of each of the regular polygons described below.

Mathematics

1 answer:

stepladder [879]2 years ago

3 0

A) 6*4 = 24cm
b) 7*3 = 21inches
c) 7*4 = 28ft
d) 9*9 = 81 inches

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Please help i am so confused

harkovskaia [24]

Answer:

$q\geq -81$

Step-by-step explanation:

First, begin by multiplying $\frac{q}{27} \\$ by its reciprocal.

To do this, flip the fraction upside-down.

The resulting equation looks like this: $\frac{27}{1} *\frac{q}{27}$

Cancel out the two 27's and you are left with $\frac{q}{1}$ , which you can simplify to q.

Next, multiply -3 by $\frac{27}{1}$ . Remember, what you do on one side, you must do on the other.

Your equation should look like this:

$q\geq -81$

The final answer is $q\geq -81$ .

Have a nice day!

3 0

3 years ago

Simplify 3x³(x+2)+2x(x²+3x)

max2010maxim [7]

Answer:

${3x}^{4} + {8x}^{3} + {6x}^{2}$

Step-by-step explanation:

${3x}^{3} (x + 2) + 2x( {x}^{2} + 3x)$

${3x}^{4} + {6x}^{3} + {2x}^{3} + {6x}^{2}$

${3x}^{4} + {8x}^{3} + {6x}^{2}$

<h3>Hope it is helpful...</h3>

8 0

3 years ago

.....................................

nirvana33 [79]

Answer:

Is this a question or..........

Step-by-step explanation:

5 0

3 years ago

Which of the following is a solution to the second order differential equation LaTeX: y''=-4y y ″ = − 4 y ? To answer this quest

Paraphin [41]

Answer:

y = sin(2t)
y = cos(2t)

Step-by-step explanation:

In the case of each of the answers listed above, the second derivative is equal to -4 times the function, as required by the differential equation.

For y = 2/3t^3, the second derivative is 4t, not -4y.

For y = e^(2t), the second derivative is 4y, not -4y.

The graph shows the sum of the second derivative and 4y is zero for the answers indicated above, and not zero for the other two proposed answers.

5 0

4 years ago

Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

<u>Answer:</u>

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

<u>Solution:</u>

Need to solve given equation using quadratic formula.

$-3 x^{2}-x-3=0$

General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

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

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$

8 0

4 years ago