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3 years ago

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Shawna, Dexter, and Tilana are solving the equation -2.5(5-n)+2=15. Shawna says, “I can begin by dividing each side of the equat

ion by –2.5 to get 5-n-2/2.5=-15/2.5." Dexter says, “I can begin by distributing -2.5 to get -12.5+2.5n+2=15." Tilana says, “I can begin by multiplying each side of the equation by -1/2,5 to get 5-n-0.8=-6." Which students are correct?

2 answers:

Salsk061 [2.6K]3 years ago

6 0

Shawna and Dexter are both correct.
I hope this helps you

laila [671]3 years ago

4 0

Hi!
Shawna is correct because since it's multiplied to the 5-n you have to do the opposite which is division.
Dexter is also correct because after distributing you can continue solving, there is nothing wrong with that.
Hope this helps!

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Answer:

$\huge\boxed{14\ \text{and}\ 7}$

Step-by-step explanation:

$n,\ m-\text{positive integer}\\\\n=2m-\text{a positive integer is twice another}\\\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}-\text{the sum of the reciprocal of the two positive integer is }\ \dfrac{3}{14}\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}n=2m&(1)\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}&(2)\end{array}\right$

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3 years ago

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erica [24]

Answer:

a = 5

Step-by-step explanation:

If it's easier for you, you can replace n(a) with y and put a as x.

Now the equation is y = 7x + 4.

It asks you to put in 39 as y.

39 = 7x + 4.

To solve this, the goal is to isolate the variable on one side. The first step I see is to subtract the 4 from both sides.

35 = 7x

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3 years ago

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balu736 [363]

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3 years ago

Find the standard equation of a sphere that has diameter with the end points given below. (3,-2,4) (7,12,4)

DiKsa [7]

Answer:

The standard equation of the sphere is $(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Let the first endpoint be represented as $(x_{1}, y_{1}, z_{1})$ and the second endpoint be $(x_{2}, y_{2}, z_{2})$ .

Hence,

Midpoint = $(\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})$

Midpoint = $(\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})$

Midpoint = $(\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\$

Midpoint = $(5, 5, 4)$

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

$d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }$

$d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}$

$d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}$

$d = \sqrt{16 +196 + 0$

$d =\sqrt{212}$

$d = 2\sqrt{53}$

This is the diameter

To find the radius, r

From $Radius = \frac{Diameter}{2}$

$Radius = \frac{2\sqrt{53} }{2}$

∴ $Radius = \sqrt{53}$

r = $\sqrt{53}$

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is $(5, 5, 4)$

Radius of the sphere is $\sqrt{53}$

The equation of a sphere of radius r and center $(h,k,l)$ is given by

$(x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}$

Hence, the equation of the sphere of radius $\sqrt{53}$ and center $(5, 5, 4)$ is

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}$

$(x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53$

This is the standard equation of the sphere

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3 years ago