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olga nikolaevna [1]

2 years ago

12

A plane traveled 580 miles to Ankara and back. the trip there was with the wind. It took 5 hours. The trip back was into the win

d. The trip back took 10 hours. Find the speed of the plane in still air and the speed of the wind.

1 answer:

Vlad1618 [11]2 years ago

3 0

290 mph , its going half the speed as before. Hope this helps.

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PLZ HELP WILL MAKR BRAINLIEST

lions [1.4K]

Answer:

It will be 0.2

Step-by-step explanation:

Hope this Helped

5 0

2 years ago

6 * ( 12 - 7.5 ) - 7

Paladinen [302]

Answer: 20

Explanation: Alright, so basically you’re going to want to use PEMDAS (Parenthesis, Exponents, Multiplication/ Division, Addition/ Subtraction)

* just means multiply.. sooo the problem looks like this:

6x(12-7.5)-7

We’ll start at the parentheses :)

So (12-7.5) = 4.5

Now our problem looks like: 6x(4.5)-7

Following PEMDAS, the next thing to come up is multiplication! So we’re going to multiply 6 by 4.5 ...

6x4.5=27

After this our problem looks like: 27-7

So now it’s just simple math...

27-7= 20

5 0

2 years ago

Read 2 more answers

Each week, Heather’s company has $5000 in fixed costs plus an additional $250 for each system produced. The company is able to p

kvv77 [185]

The question is an illustration of composite functions.

Functions c(n) and h(n) are $\mathbf{c(n) = 5000 + 250n}$ and $\mathbf{n(h) = 5h}$
The composite function c(n(h)) is $\mathbf{c(n(h)) = 5000 + 1250h}$
The value of c(n(100)) is $\mathbf{c(n(100)) = 130000}$
The interpretation is: <em>"the cost of working for 100 hours is $130000"</em>

The given parameters are:

$5000 in fixed costs plus an additional $250
5 systems in one hour of production

<u>(a) Functions c(n) and n(h)</u>

Let the number of system be n, and h be the number of hours

So, the cost function (c(n)) is:

$\mathbf{c(n) = Fixed + Additional \times n}$

This gives

$\mathbf{c(n) = 5000 + 250 \times n}$

$\mathbf{c(n) = 5000 + 250n}$

The function for number of systems is:

$\mathbf{n(h) = 5 \times h}$

$\mathbf{n(h) = 5h}$

<u>(b) Function c(n(h))</u>

In (a), we have:

$\mathbf{c(n) = 5000 + 250n}$

$\mathbf{n(h) = 5h}$

Substitute n(h) for n in $\mathbf{c(n) = 5000 + 250n}$

$\mathbf{c(n(h)) = 5000 + 250n(h)}$

Substitute $\mathbf{n(h) = 5h}$

$\mathbf{c(n(h)) = 5000 + 250 \times 5h}$

$\mathbf{c(n(h)) = 5000 + 1250h}$

<u>(c) Find c(n(100))</u>

c(n(100)) means that h = 100.

So, we have:

$\mathbf{c(n(100)) = 5000 + 1250 \times 100}$

$\mathbf{c(n(100)) = 5000 + 125000}$

$\mathbf{c(n(100)) = 130000}$

<u>(d) Interpret (c)</u>

In (c), we have: $\mathbf{c(n(100)) = 130000}$

It means that:

The cost of working for 100 hours is $130000

Read more about composite functions at:

brainly.com/question/10830110

5 0

3 years ago

colin climbs 16 feet down into tunnel and lands on the tunnel floor. then he jumps to a platform that is 2 feet above the tunnel

Ksivusya [100]

Answer:

-14

Step-by-step explanation:

-16 + 2=-14

7 0

2 years ago

Help I don’t understand at all

zloy xaker [14]

Answer:

$1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4$

Step-by-step explanation:

1.

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

$2h^2-7h+2h^2-h+6+4h-9h$

$(2h^2+2h^2)+(-7h-h+4h-9h)+(6)$

$4h^2-13h+6$

2.

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

$8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5$

$(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)$

$7x^3y^2-x^2y+1$

3.

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ( $a(b+c)=(a)(b)+(a)(c)=ab+ac$ ). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,

$-2(8n+1)-(5-9n)+3^2$

$-2(8n+1)-(5-9n)+(3*3)$

$-2(8n+1)-(5-9n)+9$

$(-2)(8n)+(-2)(1)+(-)(5)+(-)(-9n)+9$

$-16n-2-5+9n+9$

$(-16n+9n)+(-2-5+9)$

$-7n+2$

4.

The same method used to solve problem (3) can be applied to this problem.

$\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-2^3$

$\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-(2)(2)(2)$

$(\frac{1}{2})(10)+(\frac{1}{2})(-8m)+(\frac{1}{2})(6m^2})+(-)(3m^2)+(-1)(4m)+(-1)(-7)-8$

$5-4m+3m^2-3m^2-4m+7-8$

$(-3m^2-3m^2)+(-4m-4m)+(5+7-8)$

$-8m+4$

8 0

2 years ago