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guajiro [1.7K]
2 years ago
14

To

Mathematics
1 answer:
Arte-miy333 [17]2 years ago
3 0

Answer:

50 + 25 + (8*5) - (7 + 4)

Step-by-step explanation:

50 + 25 + (8*5) - 7 + 4 = 50 + 25 + 40 -  7  + 4

                                      = 115 - 7 + 4

                                      = 108 + 4

                                     = 112

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The cost of renting a bus for a field trip was split evenly among 20 students. At the last minute, 10 more students joined the t
dexar [7]

By solving a linear equation, we will see that the total cost for renting the bus is $90.

<h3>What was the total cost of renting the bus, in dollars?</h3>

Let's say that the total cost is C.

When there are 20 students, each student should pay:

p = C/20

When the other 10 students are added (for a total of 30) each student pays:

p' = C/30.

We know that the cost for each of the original 20 students decreased by $1.50, so:

p' = p - $1.50

Then we have 3 equations to work with:

p = C/20

p' = C/30.

p' = p - $1.50

Now we can replace the first and second equations into the third one:

C/30 = C/20 - $1.50

Now we can solve this linear equation for C:

C/20 - C/30 = $1.50

C*( 1/20 - 1/30) = $1.50

C*(30/600 - 20/600) = $1.50

C*(10/600) = $1.50

C*(1/60) = $1.50

C = 60*$1.50 = $90

So the total cost for renting the bus is $90.

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

4 0
1 year ago
6 pounds of apples and 3 pounds of oranges cost 24 dollars.5 pounds of apples and 4 pounds of oranges cost 23 dollars. Set up a
vitfil [10]

Answer:cost of each pound of apple= $3

And cost of each pound of orange =$2

Step-by-step explanation:

Step 1

Let cost of apples = x

And cost of Oranges =y

Let 6 pounds of apples and 3 pounds of oranges cost 24 dollars be represented as

6 x + 3y= 24----- equation 1

Also, Let 5 pounds of apples and 4 pounds of oranges cost 23 dollars be represented as

5x+ 4y= 23----- equation 2

Step 2

6 x + 3y= 24----- equation 1

5x+ 4y= 23----- equation 2

Using substitution method to solve the equation

6 x + 3y= 24

24-6x=3y

y= 24-6x/3 = 8-2x

Y= 8-2x

Substituting the value of y= 8-2x  into equation 2

5x+ 4( 8-2x)= 23

5x+ 32 -8x= 23

32-23= 8x-5x

9=3x

x=9/3

x=3

Putting the value of x= 3 in equation 1 and solving to find y

6 x + 3y= 24

6(3) +3y= 24

18+3y=24

3y= 24-18

3y=6

y=6/3= 2

Therefore the cost of each pound of apple= $3

And cost of each pound of orange =$2

8 0
3 years ago
What is the answer of a ?
Scorpion4ik [409]

Answer:

<h2>2.2</h2>

Step-by-step explanation:

Use the cosine law (look at the picture).

We have:

a=a\\b=4\\c=3\\\gamma=32^o

a^2=4^2+3^2-2(4)(3)\cos32^o

\cos32^o\approx0.848 → look at the second picture

a^2=16+9-24(0.848)\\\\a^2=25-20.352\\\\a^2=4.648\to a=\sqrt{4.648}\\\\a\approx2.2

4 0
3 years ago
(x +y)^5<br> Complete the polynomial operation
Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

\left(x+y\right)^5

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=x,\:\:b=y

=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i

so expanding summation

=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5

solving

\frac{5!}{0!\left(5-0\right)!}x^5y^0

=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5

=1\cdot \:1\cdot \:x^5

=x^5

also solving

=\frac{5!}{1!\left(5-1\right)!}x^4y

=\frac{5}{1!}x^4y

=\frac{5}{1!}x^4y

=\frac{5x^4y}{1}

=\frac{5x^4y}{1}

=5x^4y

similarly, the result of the remaining terms can be solved such as

\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2

\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3

\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4

\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5

so substituting all the solved results in the expression

=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5

=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

Therefore,

\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

6 0
2 years ago
Which expression should you multiply the numerator and denominater of
il63 [147K]

Answer: The third one

Step-by-step explanation: I just did the question on edge. ,and I got it right.

6 0
3 years ago
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