All categories

3 years ago

How many solutions does 3x+4=3x-2 have?

Mathematics

2 answers:

Burka [1]3 years ago

5 0

Answer:

No solutions.

Step-by-step explanation:

3x+4−3x=3x−2−3x

4=−2

4−4=−2−4

0=−6

0 does not equal -6, so there are no solutions.

Lera25 [3.4K]3 years ago

4 0

Answer: 2

Step-by-step explanation:

You might be interested in

solve this using pythagorean theorem please and give me the right answer and i will give YOU brainliest and a like in return.

Dima020 [189]

Answer:

6.8 miles

Step-by-step explanation:

a^2+b^2=c^2

4 0

2 years ago

Can someone please help

andrey2020 [161]

There’s no question

5 0

3 years ago

Read 2 more answers

A rectangular storage container with an open top is to have a volume of 10 m3 . then length of its base is twice the width. mate

katrin [286]

Answer:

The cost of materials for the cheapest such container is $163.54.

Step-by-step explanation:

A rectangular storage container with an open top is to have a volume of 10 m³.

The volume of the rectangle is

$\text{Volume} =\text{Length} \times \text{Width} \times \text{Height}$

Length of its base is twice the width.

Let Width be 'w'.

Length is l=2w.

Height be 'h'.

$10 =2w\times w\times h$

$10=2w^2h$

The height in terms of width is represented as,

$h=\frac{10}{2w^2}$

$h=\frac{5}{w^2}$

According to question,

The cost is 10 times the area of the base and 6 times the total area of the sides.

i.e. Cost is given by,

$C=10(L\times W)+6(2\times L\times H+2\times W\times H)$

$C=10(2w\times w)+6(2\times 2w\times \frac{5}{w^2}+2\times w\times \frac{5}{w^2})$

$C=20w^2+\frac{120}{w}+\frac{60}{w}$

$C(w)=20w^2+\frac{180}{w}$

To get the minimum value,

Differentiate the cost w.r.t 'w',

$C'(w)=20\frac{d(w^2)}{dw}+180\frac{d(w^{-1})}{dw}$

$C'(w)=20\times 2w-180 w^{-2}$

$C'(w)=40w-\frac{180}{w^2}$

To find critical points put derivate =0,

$40w-\frac{180}{w^2}=0$

$40w=\frac{180}{w^2}$

$w^3=\frac{180}{40}$

$w=\sqrt[3]{4.5}$

$w=1.65$

We find the second derivative to minimize,

$C''(w)=40\frac{d(w)}{dw}-180\frac{d(w^{-2})}{dw}$

$C''(w)=40+360(w^{-3})$

$C''(w)>0$

As $C''(w)>0$ it is the minimum cost.

The cost is minimum at w=1.65.

Substitute the values in the cost function,

$C(1.65)=20(1.65)^2+\frac{180}{1.65}$

$C(1.65)=54.45+109.09$

$C(1.65)=163.54$

Therefore, the cost of materials for the cheapest such container is $163.54.

7 0

4 years ago

0.5+(-3/4)+0.25-(-4/5)

earnstyle [38]

Answer:

th first one

Step-by-step explanationor fully:

5 0

4 years ago

Would anyone help With these questions it's the last one I don't know how to do them 1,2,3,4,5 and 6

Shtirlitz [24]

Answer:

See below

Step-by-step explanation:

1. Use x to designate the number of hours worked.

Plumber 1 = 35(x) + 25

Plumber 2 = 40(x)

Now you need to find when these two equations are equal to each other.

35(x) + 25 = 40(x)

Solve for x.

25 = 40x - 35x

25 = 5x

5 = x

So when the service call equals 5 hours, both plumbers will charge the same amount of money.

2. Use x to designate the number of months.

Center 1 = 30x

Center 2 = 22x + 80 initiation fee

Now you need to make the two equations equal to each other and solve for x.

30x = 22x + 80

30x - 22x = 80 or 8x = 80

x = 10 months

3. Use x to designate the pounds of the package

Shipper 1 = 14 + 2x

Shipper 2 = 20 + 1.5x

Make the two equations equal to each other.

14 + 2x = 20 + 1.5x

2x - 1.5x = 20 - 14

0.5x = 6

x = 12 pounds

4. Deanna $15 using $0.75 per game. This means she can play a total of 20 games before having $0.

Lise $13 using $0.50 per game. This means she can play a total of 26 games before having $0.

So since both end up with $0, you can see that Deanna would play 20 games and Lise would play 26 games.

5. Use x to designate the call time length.

Hotel 1 = 1 + 0.80x

Hotel 2 = 2 + 0.75x

Make both equations equal to each other.

1 + 0.8x = 2 + 0.75x

0.8x - 0.75x = 2 - 1

0.05x = 1

x = 20

6. Use x to designate semesters.

Duke = 22 + 6x

Kila = 4 + 12x

22 + 6x = 4 + 12x

22 - 4 = 12x - 6x

18 = 6x

3 = x semesters

6 0

3 years ago