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alexandr402 [8]
3 years ago
10

A football team lost 4 yards on each of 2 plays, gained 14 yards on the third play, and lost 5 yards on the fourth play. Write a

nd find the value of an integer expression to find the change in their field position.
Mathematics
1 answer:
Rzqust [24]3 years ago
3 0
The change in field position is one
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What are the values of x and y if this equation is true?<br><br> 22(x + yi) + (28 + 4i) 72 – 62i
ASHA 777 [7]

Answer:

x = 2

y = -3

Step-by-step explanation:

The given equation is,

22(x + yi) + (28 + 4i) = 72 - 62i

By solving this equation further,

22x + 22yi + 28 + 4i = 72 - 62i

(22x + 28) + (22y + 4)i = 72 - 62i

Now both the sides of the equation is in the form of complex number,

By comparing real and imaginary parts given on both the sides,

22x + 28 = 72

22x = 72 - 28

22x = 44

x = 2

22y + 4 = -62

22y = -62 - 4

22y = -66

y = -3

Therefore, x = 2 and y = -3 are the values for which the given equation is true.

8 0
3 years ago
You are selling raffle tickets to raise money. Each ticket costs $5. Which equation solves for the number of tickets you must se
aleksley [76]

D. 5x = 45 because you are trying to find x, the number of tickets that must be sold to earn $45.

8 0
2 years ago
If a can is 20cm Haigh X 8cm diameter what is the volume
sattari [20]
Radius=8/2=4cm

volume=4*4*3.14*20=1004.8

The answer is  1004.8cm₃

7 0
3 years ago
Solve the following matrix equations: (matrices)
Masja [62]

Step-by-step explanation:

a)

3X + \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix}  - 1 & 6 \\ 10 & 14 \end{pmatrix} \\  \\  3X  = \begin{pmatrix}  - 1 & 6 \\ 10 & 14 \end{pmatrix} -  \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}  \\  \\ 3X  = \begin{pmatrix}  - 1 - 2 & 6 - 3 \\ 10 - 4 & 14 - 5 \end{pmatrix}\\  \\ 3X  = \begin{pmatrix}   - 3 & 3 \\ 6 & 9\end{pmatrix}\\  \\ X  =  \frac{1}{3} \begin{pmatrix}   - 3 & 3 \\ 6 & 9\end{pmatrix}\\  \\ X  =  \begin{pmatrix}  \frac{ - 3}{3}  & \frac{3}{3}  \\  \\ \frac{6}{3}  & \frac{9}{3} \end{pmatrix}\\  \\ \huge \red{ X}  =  \purple{ \begin{pmatrix}  - 1  &1  \\ 2 & 3 \end{pmatrix}}

b)

3X + 2I_3=\begin{pmatrix} 5 & 0 & -3 \\6 & 5 & 0\\ 9 & 6 & 5\end{pmatrix} \\\\3X + 2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\6 & 5 & 0\\ 9 & 6 & 5 \end{pmatrix} \\\\3X + \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix} =\begin{pmatrix} 5 & 0 & - 3\\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} \\\\3X  =\begin{pmatrix} 5 & 0 & -3 \\ 6 & 5 & 0 \\ 9 & 6 & 5 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \\\\3X  =\begin{pmatrix} 5-2 & 0-0 & -3-0 \\ 6-0 & 5-2 & 0-0 \\ 9-0 & 6-0 & 5-2 \end{pmatrix} \\\\3X  =\begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X  =\frac{1}{3} \begin{pmatrix} 3 & 0 & - 3 \\ 6 & 3 & 0 \\ 9 & 6 & 3 \end{pmatrix} \\\\X  =\begin{pmatrix} \frac{3}{3}  & \frac{0}{3}  & \frac{-3}{3} \\\\ \frac{6}{3}  & \frac{3}{3}  & \frac{0}{3} \\\\ \frac{9}{3}  & \frac{6}{3}  & \frac{3}{3} \end{pmatrix} \\\\\huge\purple {X} =\orange{\begin{pmatrix} 1  & 0 & - 1\\ 2  & 1 & 0 \\ 3  & 2  & 1 \end{pmatrix}}\\

8 0
3 years ago
If Earth was 10 times farther away from the Sun than it is now, which planet would it be closest to? Compare Earth's new distanc
Phoenix [80]
You need the radius of all the orbits
earth radius is 150 million km
1.  150 million x 10 = 1500 million
From the website below Saturn would be the closest at 1,429 million km
1,500,000,000 - 1,429,000,000 = 71,000,000 km = 7.1 * 10^7 km
I hope this helps :D
8 0
3 years ago
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