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Andrews [41]
3 years ago
11

Find the slope: (-2,5) and (2,1

Mathematics
1 answer:
larisa [96]3 years ago
7 0
-1,4 I think so, good luck
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There are 3 3/4 pounds of bricks in a bag. Each brick weighs 5/8 of a pound. How many bricks are in the bag.
Bess [88]

Answer: There are 6 bricks in the bag.

Step-by-step explanation:

Convert from mixed number to decimal number. To do it, divide the numerator of the fraction by the denominator and add the result to the whole number part. Then:

3\frac{3}{4}\ lb=(3+0.75)\ lb=3.75\ lb

Convert from fraction to decimal number. To do it you need to divide the numerator by the denominator. Then, you get:

\frac{5}{8}lb=0.625\ lb

Let be "x" the number of bricks in the bag<em>.</em>

<em> </em>Based on the information given in the exercise, you can set up the following proportion:

\frac{1}{0.625}=\frac{x}{3.75}

Finally, you must solve for "x" in order to find its value. This is:

(3.75)(\frac{1}{0.625})=x\\\\x=6

8 0
3 years ago
Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)
Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

x^{2} +y^{2}+4x-1=0

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

(x-h)^2+(y-k)^2=r^2

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

      (\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2}  )

    (-2,0)

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}

r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}

r=\sqrt{5}

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius r=\sqrt{5}

The standard form of circle equation

(x-h)^2+(y-k)^2=r^2

(x-(-2))^2+(y-0)^2=\sqrt{5} ^2

on simplification is

x^{2} +y^{2}+4 x-1=0

5 0
3 years ago
345.2-126.4+712.4x634.4
scoundrel [369]
452,165.36 ? That should be the answer lol
8 0
3 years ago
What are the zeros of the function<br><img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%202x%20%5E%7B2%7D%20-8x%20-%2010" id="Tex
allsm [11]
Hello there!

To find the zeroes or the roots of a function, you just need to set the function equal to 0 then solve for x

f(x) = 2x² - 8x - 10

Set function equal to 0

2x² - 8x - 10 = 0

Now we can factorize the left side

2(x + 1)(x - 5) = 0

Then set factors equal to 0

x + 1 = 0 or x - 5 = 0

x = 0 - 1 or x = 0 + 5

x = -1 or x = 5

Thus,

The roots are -1 and 5

Let me know if you have any questions. As always, it is my pleasure to help students like you~!
7 0
3 years ago
Write an equation of the line that has a slope of 3 and contains the point (2,5) in point-slope for
enot [183]
The equation would be y=3x-1

Point Slope Formula: 

y - y1 = m(x - x1)

y - 5 = 3(x-2)
y - 5 = 3x - 6
  + 5        + 5
--------------------
y = 3x - 1


8 0
3 years ago
Read 2 more answers
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