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2 years ago

How do i solve this answer ugh please help

Mathematics

1 answer:

babunello [35]2 years ago

6 0

Separate the quadratic functions into two individual function in x which multiply together to equal the original quadratic.
Eg: x^2-3x-10=0 can be written as (x+2)(x-5)=0
The two solutions for x can then be determined by making each bracket equal to zero.
x+2=0 so x=-2x-5=0 so x=5

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3 years ago

A lamina with constant density rho(x, y) = rho occupies the given region. Find the moments of inertia Ix and Iy and the radii of

jenyasd209 [6]

Answer:

Ix = Iy = $\frac{ρπR^{4} }{16}$

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Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

Mass of disk = ρπR2

Moment of inertia about its perpendicular axis is $\frac{MR^{2} }{2}$ . Moment of inertia of quarter disk about its perpendicular is $\frac{MR^{2} }{8}$ .

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3 0

3 years ago

How many kilograms of tin will be required to make 250 kg of an alloy, if the alloy must contain 81% copper, 14% tin and 5% lead

uysha [10]

Answer:

Step-by-step explanation:

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5 0

2 years ago

Read 2 more answers

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3 years ago

You fly a hot air balloon 3.5 miles above the ground. What is the measure of BD⌢, the portion of Earth that you can see? Round y

STatiana [176]

Answer:

$\Delta s \approx 334.545\,m$

Step-by-step explanation:

Let assume that Earth is a sphere, the following trigonometric diagram is constructed and presented below. The central angle is given by this inverse trigonometric equation:

$\theta = 2\cdot \cos^{-1} \left(\frac{r}{r+h} \right)$

$\theta = 2\cdot \cos^{-1}\left(\frac{4000\,mi}{4000\,mi+3.5\,mi} \right)$

$\theta = 4.792^{\circ}$

The distance of the portion of Earth that can be seen is:

$\Delta s = \frac{4.792^{\circ}}{360^{\circ}}\cdot (2\pi)\cdot (4000\,mi)$

$\Delta s \approx 334.545\,m$

7 0

3 years ago