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4 years ago

Enter the terms and the coefficients of the expression.

Mathematics

1 answer:

Colt1911 [192]4 years ago

5 0

Answer:

Terms are 2x, -50y and -80

Coefficients are 2 and -50

Step-by-step explanation:

Given expression

2x - 50y - 80

There are three terms in the expression which includes:

A. 2x,

B. -50y and

C. -80

The coefficient of the expression includes :

2 which is the coefficient of x

-50 which is the coefficient of y

Terms are 2x, -50y and -80

Coefficients are 2 and -50

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95°<br> 6rº + 47°<br> r_ degrees

e-lub [12.9K]

Answer:

r = 8°

Step-by-step explanation:

95° = 6r° + 47° ( vertically opposite angles are equal)

6r° = 95° - 47°

6r = 48°

r = 48°/6

r = 8°

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3 years ago

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Kadeem earns a salary of $1,800 per month plus a 10% commission on his sales. In order to save money for the down payment on a h

docker41 [41]

I'm guessing 99.65
1800+10%-20%-25%=99.65
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3 years ago

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The second of two numbers is 4 less than twice the first. Their sum is 80.Find the numbers

Otrada [13]

The first number is x
the second number is 2x-4
combine both numbers = 80
x+2x-4=80
3x-4=80
3x=84
x=28
one number is 28 and the other is 52.

4 0

3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago

Please help fast need to get this and 6 more done

lara [203]

Answer:

B = 4×4×ㅈ

l = 5 cm

SA= 4×5×ㅈ + 16 ㅈ

SA = 36 ㅈ

3 0

3 years ago