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dsp73
3 years ago
8

Define the following terms

Mathematics
1 answer:
Gennadij [26K]3 years ago
5 0

Answer:

Dividend- The number that is divided is called the dividend

Divisor- a number that divides a integer ( It is the number that is divided by the Dividend)

Quotient- is the result when dividing.

Remainder-  is the amount that is left over after dividing

Step-by-step explanation:

Example

36 ÷ 6 = 6

36 is the Dividend

6 is the Divisor

6 is the Quotient

No remainder

Hopes this helps!

You might be interested in
Struggling lol. Any help would be appreciated!
Alekssandra [29.7K]

Answer:

Heptagon = 84 ft²

Octagon  = 186 yd²

Triangle = 48√3 m²

Step-by-step explanation:

\textsf{area of a regular polygon}=\dfrac12ap

where:

  • a = length of apothem
  • p = perimeter

<u>Heptagon</u>

a = 5.0 ft

p = 33.6 ft

⇒ area = 1/2 x 5.0 x 33.6 = 84 ft²

<u>Octagon</u>

a = 7.5 yd

p = 8 x 6.2 = 49.6 yd

⇒ area = 1/2 x 7.5 x 49.6 = 186 yd²

<u>Triangle</u>

Formula for the apothem of an equilateral triangle :

a=\dfrac{s}{2\sqrt{3}}  where s = side length

a = 4, so:

4=\dfrac{s}{2\sqrt{3}}

\implies 8\sqrt{3}=s

Now we know the side length, we can calculate the perimeter (p):

p = 3 x 8√3 = 24√3

⇒ area = 1/2 x 4 x 24√3 = 48√3 m²

8 0
2 years ago
What's the area below?
Marina86 [1]
I honestly have no idea :) :S ur welcome
8 0
3 years ago
Y" - 4y = (x2 - 3) sin 2x
Zolol [24]
y''-4y=0

has characteristic equation

r^2-4=0

which has roots at r=\pm2, giving the characteristic solution

y_c=C_1e^{2x}+C_2e^{-2x}

For the nonhomogeneous part of the ODE, let y_p=(a_2x^2+a_1x+a_0)\sin2x+(b_2x^2+b_1x+b_0)\cos2x. Then

{y_p}''=(-4b_2x^2+(8a_2-b_1)x+4a_1-4b_0+2b_2)\cos2x+(-4a_2x^2+(-4a_1-8b_2)x-4a_0+2a_2-4b_1)\sin2x

Substituting into the ODE gives

(-8b_2x^2+(8a_2-b_1)x+4a_1-8b_0+2b_2)\cos2x+(-8a_2x^2+(-8a_1-8b_2)x-8a_0+2a_2-4b_1)\sin2x=(x^2-3)\sin2x

It follows that

\begin{cases}-8b_2=0\\8a_2-8b_1=0\\4a_1-8b_0+2b_2=0\\-8a_2=1\\-8a_1-8b_2=0\\-8a_0+2a_2-4b_1=-3\end{cases}\implies\begin{cases}a_2=-\dfrac18\\\\a_1=0\\\\a_0=\dfrac{13}{32}\\\\b_2=0\\\\b_1=-\dfrac18\\\\b_0=0\end{cases}

which yields the particular solution

y_p=-\dfrac18x^2\sin2x+\dfrac{13}{32}\sin2x-\dfrac18x\cos2x

So the general solution is

y=y_c+y_p
y=C_1e^{2x}+C_2e^{-2x}-\dfrac18x^2\sin2x+\dfrac{13}{32}\sin2x-\dfrac18x\cos2x
4 0
3 years ago
REALLY NEED HELP: given that (-5,-2) is on the graph of f(x) find the corresponding point for the function 2f(x)
laiz [17]

we are given

point for f(x) is (-5,-2)

Whenever we need to find point for a*f(x) , we will multiply a by y-value

Here , we have 2f(x)

so, we will multiply 2 units to y-value

y-value is -2

new y-value will be 2*-2=-4

so,

point for 2f(x) will be (-5,-4)........Answer

6 0
3 years ago
Read 2 more answers
Question 2:
Anna [14]

f(x) = x^2 and g(x) = x - 3.

To find f(g(x)) replace the x in f(x) by g(x).

f(g(x)) = (x - 3)^2

= x^2 - 6x + 9.

4 0
3 years ago
Read 2 more answers
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