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madam [21]
3 years ago
6

the equation of an ellipse is x^2/100+y^2/36=1.The ellipse is centered at the origin.Find the Vertices Of the major axis and det

ermine whether the ellipse has a horizontal or vertical major axis.
Mathematics
1 answer:
Anastasy [175]3 years ago
6 0

Answer: Use the standard form to find the vertices.

Vertex

1:(10,0) vertex 2:(-10,0)

Step-by-step explanation:

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They're parallel because they have the same slope.
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What is the product of<br> (5^-4)(5^-3)
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option one is the correct answer

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The given net pay is not correct because the medicare is not 1.45% of her gross pay. The medicare should be $11.02, making the correct net pay $641.86

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3 0
3 years ago
I need help getting the answer for the 1st question
Karo-lina-s [1.5K]
If we have two similar triangles:
Triangle 1 (base 1 , height1).
Triangle 2 (base 2, height 2)
Then:

base 1 /height 1=base 2 /height 2

Data:
base 1=0.2 m
height 1= 1 m
base 2= 8 m
height 2=x

We calculate the height of the tower:
base 1 /height 1=base 2 /height 2
0.2 m / 1m=8 m / x
x=(8 m * 1 m) / 0.2 m
x=8 m²/0.2 m
x=40 m

Answer. the heigth of the tower will 40 m

3 0
3 years ago
Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t
dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: (x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14

<u>Area of the big rectangle</u>: (x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99

The difference is: 11x+85

( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

f(d)=h=-2d^2+7d+4

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

f(0)=h=-2(0)^2+7(0)+4

h=4

The height is 4 meters.

b) 9 meters.

For d=1

f(1)=h=-2(1)^2+7(1)+4

f(1)=h=-2+7+4

h=9

b) The ball travels 4 meters.

But to calculate it, it is when h=0

0=-2d^2+7d+4

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

V=\pi r^2h

V=\pi (2.5)^2 \cdot 5

V=31.25 \pi

V_{c} \approx 98.17 \text{ m}^3

The volume of the sphere:

$V=\frac{4}{3}  \pi r^2$

V_{s} \approx 65.4 \text{ m}^3

The volume of the capsule is approximately 163.57  \text{ m}^3

3 0
4 years ago
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