Ask question

Ask question

All categories

mihalych1998 [28]

2 years ago

7

when Apple sells their iPads they increase the price $50 from what it cost then to actually make the iPads one Apple Store sold

10 iPads one day which cost a total of 5000 how much does an iPad cost to exactly make?

1 answer:

Anvisha [2.4K]2 years ago

8 0

First things you have to do is figure out how much they increased the iPads by:

$50 • 10 = $500

They increased the iPad price by $500.

Subtract the total amount from the iPad price:

$5,000 - $500 = $4,500

The actual price of 10 iPads, without an increase is $4,500

Now divide that by 10 to get the amount for one iPad:

$4,500 ÷ 10 = $450

One iPad actually costs $450 to make.

You might be interested in

Two bowls and one cup weigh 800 grams. One bowl and two cups weigh 700 grams. Find the weight of the bowl and the cup

antoniya [11.8K]

Answer:

\large \boxed{\textbf{Bowl = 300 g; cup = 200 g}}

Step-by-step explanation:

Let b = the mass of a bowl

and c = the mass of a cup

We have two conditions:

$\begin{array}{lrcl}(1) &2b + c & = & 800\\(2) & b + 2c & = & 700\\\end{array}$

Calculations:

$\begin{array}{lrcll}(3) & 4b + 2c &= & 1600&\text{ Multiplied (1) by 2}\\& 3b & = & 900&\text{Subtracted (2) from (3)}\\(4) & b & = & \mathbf{300}&\text{Divided each side by 3}\\ & 600 + c & = & 800&\text{Substituted (4) into (2}\\ & c & = & \mathbf{200}\\\end{array}\\\text{The mass of a bowl is $\large \boxed{\textbf{300 g}}$ and that of a cup is $\large \boxed{\textbf{200 g}}$}$

Check:

$\begin{array}{rlcrl}2(300) + 200& =800 & & 300 + 2(200) & = 700\\600 + 200 & = 800& & 300 + 400 & = 700\\800 & = 800& & 700 & = 700\\\end{array}$

OK.

6 0

2 years ago

Seven years ago, kodi found a box of old baseball cards in the garage. since then, he has added a consistent number of cards to

Katena32 [7]

(3,52)(7,108)
slope = (108 - 52) / (7 - 3) = 56/4 = 14

y = mx + b
slope(m) = 14
(3,52)...x = 3 and y = 52
sub and find b, the y int (the original amount of cards)
52 = 3(14) + b
52 = 42 + b
52 - 42 = b
10 = b

so ur equation is y = 14x + 10....with x being the number of years and y being the total cards. <== ur equation is y = 14x + 10

He started with 10 cards....and has been adding 14 cards every year.

so after 10 years...
y = 14(10) + 10
y = 140 + 10
y = 150 <== after 10 years, he will have 150 cards

8 0

3 years ago

How are problems like these solved

Flura [38]

the angles on the opposite sides of a quadrilateral touching the circumference adds up to 180 degrees

so,

(21x-2) + (38x +5) = 180

21x+38x −2+5=180

59x+3=180

59x=177

x=3

4 0

2 years ago

Please help___________

kramer

Answer:

Step-by-step explanation:

Let's do the obvious things first.

Right rectangle

w = 11

L = 17

Formula

Area = L * W

Area = 11 * 17

Area = 187

Left rectangle

W = 8

L = 11

Area = 8 * 11= 88

Middle Rectangle

W = 11

L = 15

Area = 11 * 15 165

Now we come to the triangles. It's not obvious what to do with them. You have to infer that one of the lets is 8 and the other is 15. You get the 8 from the width of the left rectangle (see above.)

There are 2 triangles

Area = 2 * 1/2 * l1 * l2

Leg 1 = 8

Leg2 = 15

Area = 2 * 1/2 (8 * 15)

Area = <u> 120 </u>

Total = 120 + 165 + 88 + 187 = 560

Unless you have an E answer that is 560, my answer does not agree with any of the given answers. I would suggest that you ask your teacher how it is done. My method is correct.

4 0

2 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