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3 years ago

Mark hopes to one day earn $10,000. Estimate how many clients Mark would need.

Mathematics

1 answer:

andrew11 [14]3 years ago

3 0

What is his average amount of money that he earns per day so then we can solve this problem.

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Problem #18

Arisa [49]

Answer:

Step-by-step explanation:

Step one:

given data

dimension of the box

lengh= 5 in

width= 2 in

height =4 in

Required

the volume of the box

the expression for the volume is

V=L*W*H

V=5*2*4

V=40 in^3

Hence the volume of sand the box can take is 40in^3

6 0

3 years ago

Solve the system of equations by graphing. <br><br> x+y=3<br> y=2x−15

Stella [2.4K]

Hey!

Hope this helps...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

x + y = 3

x = 3 - y

y = 2x - 15

y = 2(3 - y) - 15

y = 6 - 2y - 15

y = -2y - 9

3y = -9

y = -3

y = 2x - 15

-3 = 2x - 15

-2x = -12

x = 6

So...

The answer is: (6, -3)

4 0

4 years ago

Read 2 more answers

A teacher teaches 8 students how to fold an origami model. Each of these students goes on to teach 8

valentina_108 [34]

Answer:

8x8= 64

64x6= 384

384 is the answer

Let me know if i helped!

8 0

2 years ago

Find the indicated term of the sequence whose nth term is given by the formula.

Kruka [31]

Answer:

a 12= − 3 n 2 − n

Step-by-step explanation:

3 0

3 years ago

natima [27]

Answer:

See explanation

Step-by-step explanation:

Let x be the number of spade shovels, y -the number of flat shovels and z - the number of square showels sold that day.

The store keeps an inventory of 80 shovels, then

x+y+z=80

The store always buy twice as many spade shovels as square, so

x=2z

The total cost of all shovels is

16x+9.60y+12.80z=1,072

a) The system of three equations is

$\left\{\begin{array}{l}x+y+z=80\\ \\x=2z\\ \\16x+9.60y+12.80z=1,072\end{array}\right.$

b) In matrix form this is

$\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}80\\0\\1,072\end{array}\right)$

c) The determinant is

$\left\|\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 16&9.60&12.80\end{array}\right\|=0-32+9.60-0+19.20-12.80=-16$

d) Find three determinants:

$\left\|\begin{array}{ccc}80&1&1\\ 0&0&-2\\ 1,072&9.60&12.80\end{array}\right\|=0-2,144+0-0+1,536-0=-608$

$\left\|\begin{array}{ccc}1&80&1\\ 1&0&-2\\ 16&1,072&12.80\end{array}\right\|=0-2,560+1,072-0+2,144-1,024=-368$

$\left\|\begin{array}{ccc}1&1&80\\ 1&0&0\\ 16&9.60&1,072\end{array}\right\|=0+0+768-0-0-1,072=-304$

So,

$x=\dfrac{-608}{-16}=38\\ \\y=\dfrac{-368}{-16}=23\\ \\z=\dfrac{-304}{-16}=19$

e) If the store doubled all prices and inventory, then the new matrix is

$\left(\begin{array}{ccc}1&1&1\\ 1&0&-2\\ 32&19.20&25.60\end{array}\right)\cdot \left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}160\\0\\2,144\end{array}\right)$

7 0

3 years ago