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scoundrel [369]
1 year ago
13

For each ordered pair (x, y), determine whether it is a solution to the inequality 4x+8y5-8. Is it a solution? Yes No O (6, – 4)

(-7,5) (-2, - 3) (-6,0) O O S ? Х X
Mathematics
1 answer:
avanturin [10]1 year ago
4 0

Given inequality is:

4x+8y\leq-8

Now for (6,-4), Put x=6 nd y=-4 in given inequality:

\begin{gathered} 4(6)+8(-4)\leq-8 \\ 24-32\leq-8 \\ -8\leq-8 \end{gathered}

So for (6,-4) this inequality is true.

For (-2,-3) put x=-2 and y=-3 in given inequality:

\begin{gathered} 4(-2)+8(-3)\leq-8 \\ -8-24\leq-8 \\ -32\leq-8 \end{gathered}

As -32 is less than -8 then this condition is also true.

simillarly you can check for other options also.

For (-7,5) Put x=-7 and y=5 in given inequality:

\begin{gathered} 4(-7)+8(5)\leq-8 \\ -28+40\leq-8 \\ 12\leq-8 \end{gathered}

As 12 is grater than -8 so this condition is false.

And for (-6,0) put x=-6 and y=0 in given inequality:

\begin{gathered} 4(-6)+8(0)\leq-8 \\ -24\leq-8 \end{gathered}

As -24 is less than -8 so it is true.

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GIVING BRAINLIEST!!!!!!<br><br><br>32+4X(16 x 1/2)-2<br><br>show your work
Yuliya22 [10]

Answer:

32x + 30

Step-by-step explanation:

32+4X(16 x 1/2)-2

32 + 4X x 8 -2 ( 16 times 1/2 = 8)

32 + 32x -2      ( 4 times 8 = 32)

32x + 30           ( 32 + -2 = 30)

20+(5 * 2/5)+3

20 + 2 + 3

22 + 3

25

4 0
3 years ago
The dimensions of a rectangle are √50a^3b^2 and √200a^3. What is the students error?
bearhunter [10]

Answer:

The student incorrectly simplified 30ab\sqrt{2a}+20a\sqrt{2a} .

Thus, option D is correct.

<u></u>

Step-by-step explanation:

The formula to determine the Perimeter of a rectangle of width w and length l is expressed as:

P = 2l + 2w

In other words, the perimeter can be determined by multiplying the length and width by 2 and adding the result.

In our case, the dimensions of a rectangle are \sqrt{50a^3b^2}  and  \sqrt{200a^3}\:\:.

Here is the student's solution:

2\sqrt{50a^3b^2}+2\sqrt{200a^3}=2\cdot 5ab\sqrt{2a}+2\cdot 10a\sqrt{2a}

                                 =10ab\sqrt{2a}+20a\sqrt{2a}

                                 =30ab\sqrt{2a}          

The student made an error in calculating 30ab\sqrt{2a}+20a\sqrt{2a} , because 30ab\sqrt{2a}+20a\sqrt{2a} are not like terms.

Hence, 30ab\sqrt{2a}+20a\sqrt{2a} can not be simplified to 30ab\sqrt{2a}

Therefore, the student incorrectly simplified 30ab\sqrt{2a}+20a\sqrt{2a} .

Thus, option D is correct.

<u></u>

<u></u>

<u>Here is the correct Solution:</u>

2\sqrt{50a^3b^2}+2\sqrt{200a^3}=2\cdot 5ab\sqrt{2a}+2\cdot 10a\sqrt{2a}

                                 =10ab\sqrt{2a}+20a\sqrt{2a}

7 0
3 years ago
Read 2 more answers
For which value of 0 is 0=-1?<br> A) pi/2<br> B)pi<br> C)3pi/2<br> D)2pi
harkovskaia [24]

Answer:

3\pi /2

Step-by-step explanation:

The function sin(x) adopts the value -1  for x = 3\pi /2.

That is when the segment pictured in red in the attached image is pointing from the center of the circle down. Recall that the sin function gives us the y-readout of the free end of the rotating segment in the unit circle.

5 0
3 years ago
11. Which regular polygon has a minimum rotation of 45° to carry the polygon onto itself?
ASHA 777 [7]

Answer:

the answer is a octagon that has rotation of 45degrees

3 0
3 years ago
What temperature is 9 degrees higher than -3 degrees?
aleksklad [387]

Answer:

6 degrees

Step-by-step explanation:

-3 + 9 is 6. To check this answer you can subtract 9 from 6

3 0
3 years ago
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