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miss Akunina [59]
3 years ago
14

Hello! So, Given the equation y = 2x - 8, what is the slope and the y-intercept? Please help with a good explanation of how you

got it! Please don't give me the answer! Just how you do it! Thank you~
Mathematics
2 answers:
QveST [7]3 years ago
4 0

Answer:

Step-by-step explanation:

Slope= x÷y

So. m= -2

For y intercept consider equation

y= 2x-8

y=mx+c where c is y intercept

Hence y intercept is 8 as slope is negative

elena55 [62]3 years ago
3 0

Answer:

m = 2 and b = -8.

Step-by-step explanation:

One of the most commonly used equations of straight lines is y = mx + b, where m is the slope and b is the y-intercept.

Comparing the given y = 2x - 8 to y = mx + b, we see that m = 2 and b = -8.

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The area of a rectangular park is 3/5 square mile. The length of the park is 7/8 mile. What is width of the park
Ira Lisetskai [31]
3/5 is 0.6
7/8 is 0.875
0.6/0.875 is 0.685714286
0.685714286 is 685714286/1000000000
(not so sure about it but that was what i came up with hope i helped)
6 0
3 years ago
HELPPPPPPP PLEASEEEEE !!!!
antiseptic1488 [7]

Answer:

4

Step-by-step explanation:

We know that all lengths in a rhombus are equal so we make the 2 equations equal to each other, then solve for x:

4x-6=5x-10

5x-4x=10-6

x=4

x=4

3 0
2 years ago
(25 points) Can someone please solve this I just need to see how its solved to understand
aleksley [76]

x = total amount of students in 8th Grade.

we know only one-thrid of the class went, so (1/3)x or x/3 went.

we also know 5 coaches went too, and that the total amount of that is 41.

\bf \stackrel{\textit{one third of all students}}{\cfrac{1}{3}x}+\stackrel{\textit{coaches}}{5}=\stackrel{\textit{total}}{41}\implies \cfrac{x}{3}+5=41\implies \cfrac{x}{3}=41-5 \\\\\\ \cfrac{x}{3}=36\implies x=3(36)\implies x=108

now, to verify, well, what do you get for (108/3) + 5?

4 0
3 years ago
What is 26,752 rounded to the nearest ten thousand​
Licemer1 [7]

Answer:

27,000 I think it's rii

5 0
3 years ago
Which description does NOT guarantee that a quadrilateral is a square?
Ivahew [28]
Let's go through the choices one by one

------------------------------------------
Choice A

If all sides are congruent, then this figure is a rhombus (by definition). If all angles are congruent, then we have a rectangle. Combine the properties of a rhombus with the properties of a rectangle and we have a square.

In terms of "algebra", you can think
rhombus+rectangle = square

Or you can draw out a venn diagram. One circle represents the set of all rhombuses; another circle represents the set of all rectangles. The overlapping region is the set of all squares. The overlapping region is inside both circles at the same time.

So we can rule out choice A. This guarantees we have a square when we want something that isn't a guarantee.

------------------------------------------
Choice B

If we had a parallelogram with perpendicular diagonals, then we can prove that we have a rhombus (all four sides congruent). However, we don't know anything about the four angles of this parallelogram. Are they congruent? We don't know. So we can't prove this figure is a rectangle. The best we can say is that it's a rhombus. It may or may not be a rectangle. There isn't enough info about the rectangle & square part.

This is why choice B is the answer. We have some info, but not enough to be guaranteed everytime.

------------------------------------------
Choice C

This is a repeat of choice A. Having "all right angles" is the same as saying "all angles congruent". This is because "right angle" is the same as saying "90 degrees". So we can rule out choice C for identical reasons as we did with choice A.

------------------------------------------
Choice D

As mentioned before in choice A, if we know that a quadrilateral is a rectangle and a rhombus at the same time, then the figure is also a square. This is always true, so we are guaranteed to have a square. We can cross choice D off the list.

------------------------------------------

Once again, the final answer is choice B


3 0
3 years ago
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