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Dima020 [189]
2 years ago
9

What is the equation of this line?

Mathematics
1 answer:
Margarita [4]2 years ago
3 0

Answer: either b or c

Step-by-step explanation:

¸.•´*¨`*•✿ ✧・゚: *✧・゚:* Jax *:・゚✧*:・゚✧✯✿•*`¨*`•.¸✯

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How do I round off to the nearest decimal place​
Marrrta [24]
First, find the digit in the place you are rounding to. Look at the digit one place to the right. If the digit is less than 5, round down. If the digit is 5 or greater, round up.
For example: round 1.86 to the nearest tenth
Look at the tenth digit (8). The number to the right of it is 6. Since it’s above 5, round the 8 up to 9. So you would get 1.9.

If it was 1.84, since 4 is less than 5, you would not round the 8, and you would get 1.8
3 0
3 years ago
Find the values of x when y=1
kirill [66]

Answer:

x=1+\sqrt{5}, 1-\sqrt{5}

Step-by-step explanation:

  1. x^{2} -2x-4=0
  2. (x-1)^{2} -1-4=0
  3. (x-1)^{2} -5=0
  4. (x-1)^{2} =5
  5. x-1=\sqrt{5} ,-\sqrt{5}
  6. x=1+\sqrt{5} , 1-\sqrt{5}

<u>FINAL ANSWER</u>

x=1+\sqrt{5} , 1-\sqrt{5}

<u><em>In Decimal Form (Rounded to 3 significant figures)</em></u>

<em>x≈-1.24, 3.24</em>

6 0
3 years ago
Read 2 more answers
Anyone here good at algebra? if so PLEASE HELP MEH ;-;
vovangra [49]
I’ll tell y’all 4 if did
7 0
3 years ago
Please help me find limit​
cluponka [151]

9514 1404 393

Answer:

  -13/11

Step-by-step explanation:

Straightforward evaluation of the expression at x=1 gives (1 -1)/(1 -1) = 0/0, an indeterminate form. So, L'Hopital's rule applies. The ratio of derivatives is ...

  \displaystyle\lim_{x\to 1}\dfrac{n}{d}=\dfrac{n'}{d'}=\left.\dfrac{\dfrac{4}{3\sqrt[3]{4x-3}}-\dfrac{7}{2\sqrt{7x-6}}}{\dfrac{5}{2\sqrt{5x-4}}-\dfrac{2}{3\sqrt[3]{2x-1}}}\right|_{x=1}=\dfrac{4/3-7/2}{5/2-2/3}=\dfrac{8-21}{15-4}\\\\=\boxed{-\dfrac{13}{11}}

4 0
3 years ago
Someone help with this
elena-14-01-66 [18.8K]

The quadrants are I, II, III, and IV, (meaning 1, 2, 3, and 4). The first quadrant is in the upper right, which has both positive x and positive y values. The second quadrant is the upper left, which has negative x and positive y values. The third quadrant is the lower left, which has both negative x and y values. The fourth quadrant is the lower right, which has positive x and negative y values. Using this knowledge and our positive and negative signs, the following are the answers to questions 1-6.

1) (-4, -2) - Quadrant III, both x and y are negative

2) (0, -7) - This point is actually on the y-axis. The x value is 0 and the y value is -7, so the graph is 7 units down from the origin on the y-axis.

3) (0,0) - This point is the origin, or where the x and y axes cross (the middle of the graph).

4) (6, -9) - This point is in Quadrant IV, because it has a positive x value and a negative y value

5) (3,5) - This point is in Quadrant I, because both the x and y values are positive.

6) (8,0) - This point is on the x-axis. The y-value is zero, so this point is 8 units to the right of the origin between Quadrant I and Quadrant IV.

Using the knowledge presented above, to graph the points given to you in the second part of the problem, first you can figure out what quadrant or part of the graph the point is on. Then, you can count the number of units (squares on the graph) in the right direction (remember that up is positive on the y-axis and down is negative, and to the right is positive on the x-axis and to the left is negative) in order to plot the points. Then, you must connect the points that correspond to the same figure in order to create the figures.

Please comment if you have any questions!

Hope this helps!

7 0
3 years ago
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