Three more than four times a number is 15

Answers

Answer 1

Answer: You can model this with the equation 4x+3=15 since it is a number (x) times 4 with 3 added and equals 15.
Subtract 3 from both sides
4x=12
Divide both sides by 4
x=3

Answer 2

Answer: 4x +3=15 is the equation ..... You do not have to solve it

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Simplify this problem 3(x + 1) -2x=

Answers

The answer is x + 3

Step 1: Expand ( 3x + 3 - 2x )

Step 2: Collect like terms ( (3x - 2x) + 3 )

Step 3: Simplify ( x + 4 )

Let me know if I helped :)

patrick read 20 pages of his book in 4 hours . Todd read 25 pages in 5 hours. Did both at the same rate?

Answers

All you need to do is find the rate that each boy read and compare them.

Patrick
20 pages/4 hours
=5 pages/hour

Todd
25 pages/5 hours
=5 pages/hour

Yes both boys read at the same rate (5 pages/hour)

Solve:-

Patrick:-
20 pgs = 4 hrs

Todd:-
25 pgs = 5 hrs

Divide:-

Patrick:-

20 ÷ 4 = 5
5 pgs per hr

Todd:-

25 ÷ 5 = 5
5 pgs oer hr

Yes, Both boys read at the same rate, 5 pgs per hr

Identify the coefficient, variable, and exponent of 3y.

Answers

Answer:

the coefficient is 3, the variable is y and the exponent is 1.

Step-by-step explanation:

the coefficient is always the number, it can stand alone, or have multiple variables attached. The variable is the letter in an equation, they also can stand alone or follow a coefficient. The exponent is not shown, but if you were to multiply y once, it would equal y.

Hope this helps.

Please please please help me!!!

Answers

if thats plato, j copy the answer and search for that, its what I do and most of the time the answers are on Brainly

Simple question

Derivative of $\boxed{f(y)= (y^2)/(y^3+8) }$

Answers

Let's go ;D

$f(y)=(y^2)/(y^3+8)$

we have to use the quotient rule.

$f(y)=(g(y))/(h(y))$

$f'(y)=(h(y)*g'(y)-g(y)*h'(y))/([h(y)]^2)$

Then

$g(y)=y^2$

$g'(y)=2y$

$h(y)=y^3+8$

$h(y)=3y^2$

Now we can replace

$f'(y)=(h(y)*g'(y)-g(y)*h'(y))/([h(y)]^2)$

$f'(y)=((y^3+8)*2y-(y^2)*3y^2)/((y^3+8)^2)$

$f'(y)=(2y^4+16y-3y^4)/((y^3+8)^2)$

$\boxed{\boxed{f'(y)=(16y-y^4)/((y^3+8)^2)}}$

Solve for p. N=P-K/J

Answers

$N = P- (K)/(J)$

$P = N+ (K)/(J)$

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Please help with this algebra :)