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4 years ago

How I solve 2x-1=10? Thx

Mathematics

2 answers:

Lelechka [254]4 years ago

7 0

Answer:

The answer is 5.5

Step-by-step explanation:

First add one to both sides giving you 2x=11

Then simply divide 11 by 2

Resulting in 5.5

Oxana [17]4 years ago

4 0

Answer:

x=5.5

Step-by-step explanation:

2x-1=10

+1 +1

2x=11

divide by 2 on each side

x=5.5

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3) Mr. Brust is addicted to Red Bull and has decided that he wants to plaster his room at school with Red Bull cans.

Bogdan [553]

Mr. Brust needs 427 cans to cover the wall of dimensions 20 feet long and 9 feet tall.

<h3>What is area of rectangle?</h3>

Area of rectangle is the product of its sides

i.e., Area of rectangle = length x breadth

<h3>What is area of circle?</h3>

The area of a circle is pi times the radius squared (A = π r²).

<h3>What is circumference of circle?</h3>

The circumference of a circle is defined as the linear distance around it.

( C= 2πr)

The can of red bull is in shape of cylinder.

If we open the cylinder it will have two circular surfaces( at bottom and top) and one rectangular sheet.

Like the given figure given.

The can is 6.5 inches tall and has a diameter of about 2.5 inches.

It shows that the two circle will have the diameter 2.5 inches.

d= 2.5 inch

= 0.208 ft

radius, r= $\frac{d}{2}$

r= $\frac{0.208}{2}$

r= 0.104 ft

To get the length of the rectangle , we have to calculate the circumference of the circle.

Circumference of the circle be,

= 2πr

= 2 x 3.14 x 0.104

= 0.653 ft

The rectangular sheet will have,

length= 0.653 ft, breadth= 6.5 inches = 0.541 ft

Now, area of rectangular sheet be,

= length x breadth

= 0.653 x 0.541

= $0.3532 ft^{2}$

Area of 2 circle( top and bottom of can) be,

= 2 x π $r^{2}$

= 2 x 3.14 x 0.104 x 0.104

= 6.28 x 0.010816

= 0.0679 $ft^{2}$

Total area of can be,

= Area of rectangular sheet + Area of 2 circle( top and bottom of can)

= 0.3532 + 0.0679

= 0.4211 $ft^{2}$

Now, we have the wall of dimensions

length= 9 ft, breadth= 20 ft

Area of wall be,

= length x breadth

=9 x 20

= 180 $ft^{2}$

Number of cans required to cover the whole wall be,

= $\frac{area \;of\; wall}{area \;of\; can}$

= $\frac{180}{0.4211}$

= 427.45 cans

≈ 428 cans

Learn more about the concept of area here:

brainly.com/question/16193772

#SPJ1

7 0

2 years ago

A chef has less than 25.8 lb of rice. She takes out 5 lb to use during the day and stores the rest in containers. Each container

klasskru [66]

25.8>2.5x+5, so 20.8=2.5x, so 8.32=x. Therefore, 8 containers will be completely full.

6 0

3 years ago

Help pleasee , thank you so much in advance

sp2606 [1]

Answer:

A - 2x + 4 + 50 = 10x - 2

Step-by-step explanation:

1.) It is the only equation that makes mathematical sense

2.) The sum of angles ABD and DBC equal angle ABC, and ABC = 10x -2

Note:

x=7 if it is important later on

8 0

3 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

3 years ago

How many positive integers $n$ satisfy $127 \equiv 7 \pmod{n}$? $n=1$ is allowed.

Svetllana [295]

Naturally, any integer $n$ larger than 127 will return $127\equiv127\mod n$ , and of course $127\equiv0\mod127$ , so we restrict the possible solutions to $1\le n$ .

Now,

$127\equiv7\mod n$

is the same as saying there exists some integer $k$ such that

$127=nk+7$

We have

$\implies 120=nk$

which means that any $n$ that satisfies the modular equivalence must be a divisor of 120, of which there are 16: $\{1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\}$ .

In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

$127=21\cdot6+1\iff127\equiv1\equiv7\mod6$

(If we're allowing $n=1$ , then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)

5 0

4 years ago