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GrogVix [38]
3 years ago
6

HHHHHHEEEELLLLLLPPPPP

Mathematics
1 answer:
sveticcg [70]3 years ago
4 0
<span>We want to check how many intersections line A and B have, that is, we want to check how many common solutions do these equations have:
</span>
i) 2x + 2y = 8

ii) x + y = 4
<span>
use equation ii) to write y in terms of x as : y=4-x, 

substitute y =4-x in equation i):

</span>2x + 2y = 8
2x + 2(4-x) = 8
<span>2x+8-2x=8
8=8

this is always true, which means the equations have infinitely many common solutions.


Answer: </span><span>There are infinitely many solutions.</span><span>

</span>
You might be interested in
A rectangular field is 75 yards wide and 105 yards long
nika2105 [10]

Answer:

width = 72 yards

length = 108 yards

Step-by-step explanation:

Given:

  • Width = 75 yards
  • Length = 105 yards

<u>Area of the field</u> with the given values:

\begin{aligned}\textsf{Area of a rectangle}&=\sf width \times length\\& = \sf 75 \times 105\\& = \sf 7875\:\: yd^2\end{aligned}

To maintain the <u>same perimeter</u>, but <u>change the area</u>, either:

  • decrease the width and increase the length by the same amount, or
  • increase the width and decrease the length by the same amount.

In geometry, length pertains to the <u>longest side</u> of the rectangle while width is the <u>shorter side</u>.  Therefore, we should choose:

  • decrease the <u>width</u> and increase the <u>length</u> by the <u>same amount</u>.

<u>Define the variables</u>:

  • Let x = the amount by which to decrease/increase the width and length.

Therefore:

\implies \sf width \times length < 7875\:\:yd^2

\implies (75-x)(105+x) < 7875

Solve the inequality:

\begin{aligned}(75-x)(105+x) & < 7875\\7875-30x-x^2 & < 7875\\-x^2-30x & < 0\\-x(x+30) & < 0\\x(x+30) & > 0\\\implies x & > 0 \:\: \textsf{ or }\:\:x < - 30\end{aligned}

Therefore, as distance is positive only and the maximum width is 75 yd (since we are subtracting from the original width):

\begin{cases}\textsf{width} = 75 - x\\\textsf{length} = 105 + x\end{cases}

\textsf{where } 0 < x < 75

Therefore, to find the width and length of another rectangular field that has the same perimeter but a smaller area than the first field, simply substitute a value of x from the restricted interval into the found expressions for width and length:

<u>Example 1</u>:

  • Let x = 3

⇒ Width = 75 - 3 = 72 yd

⇒ Length = 105 + 3 = 108 yd

⇒ Perimeter = 2(72 + 108) = 360 yd

⇒ Area = 72 × 108 = 7776 yd²

<u>Example 2</u>:

  • Let x = 74

⇒ Width = 75 - 74 = 1 yd

⇒ Length = 105 + 74 = 179 yd

⇒ Perimeter = 2(1 + 179) = 360 yd

⇒ Area = 1 × 179 = 179 yd²

4 0
1 year ago
Read 2 more answers
Solve -8x + 4 &lt; 36.<br><br>A. x &gt; 4<br>B. x &lt; 4<br>C. x &lt; -4<br>D. x &gt; -4​
andre [41]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

let's solve for x ~

  • - 8x + 4 < 36

  • - 8x < 36 - 4

  • - 8x < 32

  • - x  < 32 \div 8

  • - x  < 4

  • x   >  - 4

6 0
3 years ago
Read 2 more answers
A telephone call costs c cents for the first 3 minutes and m cents for each additional minute. Which expression represents the c
tekilochka [14]

Answer: c+3m

Step-by-step explanation:

Given

A telephone call costs c cents for first 3 minutes and m cents for each additional minute.

Any call beyond 3 minutes costs

\Rightarrow c+mt

where t is the minutes spent beyond 3 minutes

For 6 minutes , t is 3

thus, it can be written

\Rightarrow c+3m

4 0
3 years ago
A circle’s diameter is 22 inches. What is the circle’s circumference
alina1380 [7]

Answer:

69.12 in, or 22pi in terms of pi

Step-by-step explanation:

Circumference of circle formula: 2piR

Radius of 22 in diameter is 11 in

So, 2pi(11) is about 69.12 in

8 0
3 years ago
Last picture was super blury
Yuri [45]
25y^8/49x^3 is the answer to this question
HOPE IT HELPS YOU
4 0
3 years ago
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