All categories

3 years ago

How do you solve for X ?4x-1=3x+3

2 answers:

5 0

You have to move like kinds to the same side. I then would be 4x-3x=3+1

Meaning your answer would be 1x=4 So x=4

Check it by replacing.. 4(4)-1 = 3(4)+3
16-1=12+3
15=15

Mazyrski [523]3 years ago

3 0

Given 4x-1=3x+3
First, get x on one side.
Do this by adding 1 to both sides.
4x-1+1=3x+3+1
You get: 4x=3x+3
Subtract 3x from both sides: x=3.

You might be interested in

I need help on the one about slope

pantera1 [17]

I think it is -5
If you divide -1/2 and then 4/1 you will get -5/1 or also known as -5

7 0

2 years ago

Day, everything at a store is on sale. The store offers a 20% discount.

Nutka1998 [239]

Answer:

$14.40

Step-by-step explanation:

$18-20%=$14.40

4 0

2 years ago

A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat

jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.

We are given a function (corrected)

$N(t) = 20(t^2-lnt^2)+ 30$

$N(t) = 20(t^2-2lnt)+ 30$

(a)

First, we take its derivative

$N'(t) = 20(2t-\frac{2}{t})$

Solve N'(t)=0

$20(2t-\frac{2}{t})=0$

Simplifying

$2t^2-2=0$

Solving for t

$t=1\ ,t=-1$

Only t=1 belongs to the valid interval $1\leqslant t\leqslant 15$

Taking the second derivative

$N''(t) = 20(2+\frac{2}{t^2})$

Which is always positive, so t=1 is a minimum

(b)

$N(1)=20(1^2-2ln1)+ 30$

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

$N(15)=20(15^2-2ln15)+ 30$

N(15)=4391.7 is a maximum

7 0

3 years ago

What is the equation of the line that passes through the points (-1, 7) and (2, 10) in Standard Form?

Usimov [2.4K]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{10-7}{2-(-1)}\implies \cfrac{3}{2+1}\implies \cfrac{3}{3}\implies 1$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=1[x-(-1)]\implies y-7=x+1 \\\\\\ y=x+8\implies \boxed{-x+y=8}\implies \stackrel{\textit{standard form}}{x-y=-8}$

just to point something out, is none of the options, however -x + y = 8, is one, though improper.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20-%203y%20%7B%7D%5E%7B2%7D%20%3D%20%20-%2075" id="TexFormula1" title=" - 3y {}^{2} = - 75"

Paraphin [41]

Answer:

The answer will turn out to be y= 5 or y= -5.

7 0

3 years ago