Solve for x, 4x – 5 = -17

Answers

Answer 1

Answer: x= -3
4(-3) = -12
-12-5= -17

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Which of the following situations could be represented by the equation y = 3x + 5?

Solve the formula A=lw for l

Answers

The solution for l in terms of A and w is: l = A / w. To solve the formula A = lw for l (length), we need to isolate the variable l on one side of the equation.

Follow these steps:

Step 1: Start with the formula: A = lw

Step 2: Divide both sides of the equation by w to isolate l:

A / w = lw / w

Step 3: Simplify the right side of the equation:

A / w = l * (w / w)

Step 4: Since w / w equals 1, we have:

A / w = l * 1

Step 5: Therefore, l is equal to:

l = A / w

So, the solution for l in terms of A and w is:

l = A / w

To know more about length:

brainly.com/question/17010889

#SPJ6

A=lw
divide boh sides by 2
A/w=l

Arrange the equations in the correct sequence to find the inverse of f(x)=y=3x/8+x

Answers

To find the inverse of the the function, it must be arranged in such a way that the independent variable be on the left side and the dependent variable on the right. So,
f(x) = y = 3x/ (8 + x)
y (8 + x) = 3x
8y + xy = 3x
8y = 3x - xy
8y = x (3 - y)
8y / (3-y) = x
x = 8y / (3 -y)

Therefore,
f-(x) = f(y) = x = 8y/ (3 - y)

Please help me I don't know how to do it

Answers

findd total collected
add everybody
fri+sat+sun+mon=
21+5 and 5/6+2.75+4 and 5/8
add what you can
21+5+2.75+4+5/6+5/8=
28.75+5/6+5/8
add the fractions by making denom same
5/6+5/8=20/24+15/24=35/24=1 and 9/24=1 and 3/8
28.75+1 and 3/8=29.75+3/8
that is how many barels total
times 205 since that is how many litesr in each bottle
205 times (29.75+3/8)=
205*29.75+205*(3/8)=
6098.75+76.875=
6175.625

29/30 is lost

1//30 remains
9175.625 times 1/30 (aka divided by 30)=205.854166666666666 litesr
205 bottles (and 0.85 partially full bottle)

Solve the following equation.2/3x + 6 = 1/2x + 1/4x

2
6
14 2/5
72

Answers

Answer: 72
heyy, umm basically group like terms together. E.g. 6= 1/2x + 1/4x - 2/3x. Then, just solve for x e.g. 1/2 + 1/4 - 2/3 and use the answer to equal it to 6. Which should be 1/12x=6... pretty sure haha. Then divide. E.g.6/1/12 for x. In this case I got 72.. :)

Find solution to logarithm.

Answers

$\log5x+\log(x-1)=2\n D:5x>0 \wedge x-1>0\n D:x>0 \wedge x>1\n D:x>1\n \log5x(x-1)=2\n 10^2=5x^2-5x\n 5x^2-5x-100=0\n x^2-x-20=0\n x^2-5x+4x-20=0\n x(x-5)+4(x-5)=0\n (x+4)(x-5)=0\n x=-4 \vee x=5\n -4\not \in D\Rightarrow x=5$

$log_pa=b\ \ \ \Leftrightarrow\ \ \ p^b=a\n\nloga+logb=log(a\cdot b)\ \ \ \Rightarrow\ \ \ D:\ a>0\ \ \ and\ \ \ b>0\n ----------------------------- \nlog(5x)+log(x-1)=2\ \ \ \Rightarrow\ \ \ D:\ 5x>0\ \ \ and\ \ \ x-1>0\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>0\ \ \ and\ \ \ \ \ \ \ \ x>1\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(1;+\infty)\n\nlog[5x\cdot(x-1)]=2\ \ \ \Leftrightarrow\ \ \ 5x(x-1)=10^2$

$5x^2-5x=100\ /:5\n\nx^2-x-20=0\n\nx^2-5x+4x-20=0\n\nx(x-5)+4(x-5)=0\n\n(x-5)(x+4)=0\ \ \ \Leftrightarrow\ \ \ x-5=0\ \ \ or\ \ \ x+4=0\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=5\ \ \ or\ \ \ \ \ \ \ \ x=-4\n\n5\in D;\ \ \ -4\notin D\n\nAns.\ x=5$

a teacher and 10 students are to be seated along a bench in the bleachers at a basketball game. In how many ways can this be done if the teacher must be seated in the middle and a difficult student must sit to the teachers immediate left?

Answers

Wow !

OK. The line-up on the bench has two "zones" ...

-- One zone, consisting of exactly two people, the teacher and the difficult student.
   Their identities don't change, and their arrangement doesn't change.

-- The other zone, consisting of the other 9 students.
   They can line up in any possible way.

How many ways can you line up 9 students ?

The first one can be any one of 9.   For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.

   The total number of possible line-ups is

   (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .

But wait ! We're not done yet !

For each possible line-up, the teacher and the difficult student can sit

-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.

That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .

So the total total number of ways to do this is

   (362,880) x (10) = 3,628,800 ways.

If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !

Answer:

Step-by-step explanation:

362,880

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