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xenn [34]
3 years ago
8

Help me maybe? Please? :))

Mathematics
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:

4 1/12

Step-by-step explanation:

7/12(7)

49/12

<em>12 goes into 49 four times. 12 x 4 is 48, which leaves 1/12 left over. </em>

4 1/12

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P and Q are two points on the line x-y+1=0 and are at distance 5 from the origin. Find the coordinates of P and Q.
Rashid [163]

Answer:

P and Q are two points on the line x-y+1=0 and are at the distance 5 units from origin. Find the area of the triangle OPQ. P and Q are two points on the line x-y+1=0 and are at the distance 5 units from origin. ... The area of PQR is 7*7/2 = 49/2 sq units.

4 0
3 years ago
Which of the following points is a solution to the system of equations shown?
Anna [14]

Answer:

(-1, 5)

Step-by-step explanation:

Rearrange the the first equation to have y on one side and all other values on the other side

y = 1 - 4x

since both of the equations are equal to y, set them equal to each other an solve for x.

1 - 4 x = x + 6

-5 = 5 x

x = -1

plug this back into the second equation to get y = 5

6 0
4 years ago
HELP PLEASE.<br> ALGEBRA.
serg [7]
the answer is a because the +2 is the why intercept.
6 0
3 years ago
Read 2 more answers
How do you solve 2z - 31 = -9z + 24?
Schach [20]

2z−31=−9z+24

Add 9z to both sides.

2z−31+9z=−9z+24+9z

11z−31=24

 Add 31 to both sides.

11z−31+31=24+31

11z=55

Divide both sides by 11.

11z/11=55/11

z=5

6 0
3 years ago
Read 2 more answers
A rectangular patio has a length that is 4 times its width and has an area of 5000 square feet. What is the width of the patio?
dimaraw [331]

\bf \underline{ \underline{Given : }}

  • A rectangular patio has a length that is 4 times its width.

  • Area of rectangular patio is 5000 square feet.

\bf \underline{ \underline{To  \: be  \: calculated :  }}

Calculate the width of the patio.

\bf \underline{ \underline{Solution : }}

Let the width be a.

Then,Length = 4a

<u>According</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>question</u><u>, </u>

<u>\sf \: Area  \: of \:  rectangle = 5000 \:  ft {}^{2}</u>

\sf \implies \: Length  \times  Width  = 5000

Substituting the values of Length and width, we get..

\sf \implies 4a \times a = 5000

\sf \implies4 {a}^{2}  = 5000

\sf\implies \:  {a}^{2}  =  \cancel \dfrac{5000}{4}

\sf\implies \:  {a}^{2}  = 1250

\sf \implies \: a =  \sqrt{1250}

\sf \implies \: a = 35.35 \:  \: (approx)

\sf\therefore \: The \: width  \: of \:  rectangular  \: patio \:  is  \: 35.35 \: ft

3 0
3 years ago
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