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Archy [21]
4 years ago
9

What rectangle shows the final image

Mathematics
1 answer:
frozen [14]4 years ago
6 0
Is there an image attached?
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A jewellery shop is having a sale.
bulgar [2K]

Answer:

f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)f(x) = x2 – p(x + 1) – c, then (α + 1)(β + 1)

Step-by-step explanation:

A jewellery shop is having a sale.

A bracelet is reduced to £240.

a) What is the original price?

A ring is reduced to £840.

This is 40% of the original price.

b) Work out the original price of this ring.

4 0
3 years ago
a tourist drove his car for the total of 17.5 hours while on a 3-day vacation on the first day he drove 4.5 hours in the second
Gelneren [198K]

Answer:

Step-by-step explanation:

34hr

6 0
3 years ago
Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for th
Masteriza [31]

Answer:

Let t_i be the time for the ith task.

We know these times have a certain structure:

  • Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5  \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6

  • The second task takes 1 second t_2=1
  • The fourth task takes 10 seconds t_4=10

So, we have the following system of equations:

t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0  \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

  • Subtract row 2 from row 1 \left(R_1=R_1-R_2\right).
  • Subtract row 2 from row 3 \left(R_3=R_3-R_2\right).
  • Add row 3 to row 2 \left(R_2=R_2+R_3\right).
  • Multiply row 3 by −1 \left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right).
  • Add row 4 multiplied by \frac{3}{2} to row 1 \left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right).
  • Subtract row 4 from row 3 \left(R_3=R_3-R_4\right).

Here is the reduced echelon form for the augmented matrix.

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]

c) The additional rows are

\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right

and the augmented matrix is

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]

d) To solve the system you must use these row operations.

  • Subtract row 1 from row 6 \left(R_6=R_6-R_1\right).
  • Subtract row 2 from row 6 \left(R_6=R_6-R_2\right).
  • Subtract row 3 from row 6 \left(R_6=R_6-R_3\right).
  • Swap rows 5 and 6.
  • Add row 5 to row 3 \left(R_3=R_3+R_5\right).
  • Multiply row 5 by 2 \left(R_5=\left(2\right)R_5\right).
  • Subtract row 6 multiplied by 1/2 from row 1 \left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right).
  • Add row 6 multiplied by 1/2 to row 3 \left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right).

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]

The solutions are: (t_1,...,t_6)=(5,1,44,10,90,20).

5 0
3 years ago
Please help me out!!!!!!
Anna71 [15]

Volume of a pyramid is (L * w * h) / 3

L=10 and w=10

h can be found by using the Pythagorean theorem:

divide the base by 2 and we get 5. Then we have that the hypotenuse is 13 and the bottom of the right triangle we formed is 5. a= sqrt(c^2-b^2)

a=h=12

we plug these numbers into the formula for the volume of a pyramid and get:

400

3 0
3 years ago
12x+6y=24 for x <br> a).x=1/2y+2<br> b).x=24-6y<br> c).x=12(24-6y)<br> d).x=2-1/2y
Nina [5.8K]

Answer:

12x+6y=24 ( dividing by 12)

x+(1/2)y=2

<u>x=2-(1/2)y</u>

<u>D</u>

<u />

5 0
3 years ago
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