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2 years ago

9) The list price of a car is $12,600.00. Liza gets a 12% discount on the list price. How much does she need to pay for the car?

Mathematics

1 answer:

o-na [289]2 years ago

8 0

Answer:

11,088

Step-by-step explanation:

12,600.00*(1-0.12)=11088

1-0.12 is the reduction

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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

6 0

3 years ago

827 divided by 1120

KiRa [710]

It can be the last one, 827 over 1120

Hope that helps

4 0

3 years ago

Read 2 more answers

^^^^<br> please help :)))

ikadub [295]

2x-1.1/4+x/3=2

2x-1/4+x/3=2

2x-1/4+x/3+1/4=2+1/4

2x+x/3=9/4

3(2x+x/3)=3.94

7x=27/4

7x/7=27/4/7

x=27/8

<h2>PLEASE MARK ME AS BRAINLIEST IF U LIKE MY ANSWER AND SORRY FOR GIVING THE ANSWER LATE BECAUSE I'VE GIVEN U ANSWER FROM MY LAPTOP PLEASE TELL THAT IT'S CORRECT OR NOT</h2>

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3 years ago

Given: ∆ABC –iso. ∆, m∠BAC = 120°

lidiya [134]

Answer: $P\triangle ADH = (a+b+\sqrt{a^2+b^2} )$ unit

Step-by-step explanation:

Since Here Δ ABC is a isosceles triangle.

Also, $AH\perp BC$ and $HD\perp AC$

Thus, ∠ ADH = 90°

That is, Δ ADH is the right angle triangle.

In which, AD = a unit and HD = b unit

And, By the definition of Pythagoras theorem,

$AH^2 = HD^2 + AD^2$

⇒ $AH^2 = a^2 +b^2$

⇒ $AH = \sqrt{a^2+b^2}$

Since, the perimeter of a triangle = sum of the all sides of the triangle.

Therefore, perimeter of ΔADH = AD + DH + AH

= $a + b + \sqrt{a^2+b^2}$ unit

4 0

3 years ago

43 ÷ (-7) i need a answer quick

vlabodo [156]

Answer:

look it up my guy

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers