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3 years ago

Can someone help me?

Mathematics

2 answers:

neonofarm [45]3 years ago

8 0

A, B and D are possible

to do that, put -3 as x and 2 as y and check whether the equation is true or not.

good luck

astra-53 [7]3 years ago

4 0

Slope m = 2/3 passing thru (-3, 2)
equation in point slope form
y - y1 = m(x - x1)
so
y - 2 = 2/3(x + 3) -------------answer is A.
y = 2/3 x + 2
y = 2/3 x + 4 --------------answer is D.
or
3y = 2x + 12
2x - 3y = -12 ------------------answer is B

Answers:
A, B and D.

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A 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find the probability of rolling

UNO [17]

1/6 is the probability of rolling a dice and getting less than 3.

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3 years ago

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

algol [13]

Find the critical points of $f(x,y)$ :

$\dfrac{\partial f}{\partial x}=-2x=0\implies x=0$

$\dfrac{\partial f}{\partial y}=y-4y^3=y(1-4y^2)=0\implies y=0\text{ or }y=\pm\dfrac12$

All three points lie within $D$ , and $f$ takes on values of

$\begin{cases}f(0,0)=4\\f\left(0,-\frac12\right)=\frac{65}{16}\\f\left(0,\frac12\right)=\frac{65}{16}\end{cases}$

Now check for extrema on the boundary of $D$ . Convert to polar coordinates:

$f(x,y)=f(\cos t,\sin t)=g(t)=4-\cos^2-\sin^4t+\dfrac12\sin^2t=3+\dfrac32\sin^2t-\sin^4t$

Find the critical points of $g(t)$ :

$\dfrac{\mathrm dg}{\mathrm dt}=3\sin t\cos t-4\sin^3t\cos t=\sin t\cos t(3-4\sin^2t)=0$

$\implies\sin t=0\text{ or }\cos t=0\text{ or }\sin t=\pm\dfrac{\sqrt3}2$

$\implies t=n\pi\text{ or }t=\dfrac{(2n+1)\pi}2\text{ or }\pm\dfrac\pi3+2n\pi$

where $n$ is any integer. There are some redundant critical points, so we'll just consider $0\le t< 2\pi$ , which gives

$t=0\text{ or }t=\dfrac\pi3\text{ or }t=\dfrac\pi2\text{ or }t=\pi\text{ or }t=\dfrac{3\pi}2\text{ or }t=\dfrac{5\pi}3$

which gives values of

$\begin{cases}g(0)=3\\g\left(\frac\pi3\right)=\frac{57}{16}\\g\left(\frac\pi2\right)=\frac72\\g(\pi)=3\\g\left(\frac{3\pi}2\right)=\frac72\\g\left(\frac{5\pi}3\right)=\frac{57}{16}\end{cases}$

So altogether, $f(x,y)$ has an absolute maximum of 65/16 at the points (0, -1/2) and (0, 1/2), and an absolute minimum of 3 at (-1, 0).

5 0

3 years ago

What is the perimeter, in units, of polygon UVWXYZ? Enter your answer in the box.

nikitadnepr [17]

Given:

Consider the below figure attached with this question.

To find:

The perimeter of the polygon UVWXYZ.

Solution:

All sides and vertical and horizontal. So, we can easily find the side lengths of the given polygon by counting the boxes between two points.

From the figure it is clear that,

UV = 9 units

VW = 2 units

WX = 3 units

XY = 3 units

YZ = 6 units

ZU = 5 units

Now, perimeter is the sum of all the sides.

$Perimeter=UV+VW+WX+XY+YZ+ZU$

$Perimeter=9+2+3+3+6+5$

$Perimeter=28$

Therefore, the perimeter of UVWXYZ is 28 units.

6 0

3 years ago

Please help right away

andrew-mc [135]

Answer:

100

Step-by-step explanation:

Mixed candy question... Skittles jar... to be filled with Jelly beans.

Let's first calculate the volume of the jar. We'll assume it's a regular cylindrical prism jar, unlike the one on the photo which is narrower on top.

V = π * r² * h = π * (3.5)² * 11.5 = 140.875 π = 442.6 cubic cm

Now, we don't have the precise measurement of a jelly bean, but we know it's roughly 2-3 cubic cm. The precision isn't needed to answer this question, just to have a rough idea... it's no 300 cu cm per jelly bean.

So, let's assume a 3 cu cm per jelly bean (2 cu cm wouldn't the final answer)....

442.6 / 3 = 147.5 jelly beans, approximately.

So, can they fit 100,000? No

Can we fit 10,000 in there? No

Can we fit 100? Yes.

Can we fit 1? Certainly

The most reasonable lower-limit would then be 100.

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3 years ago

Find the ratio a:b- 1) 2a=9b 2) a+b=3b please explain your work

Alex777 [14]

1) 2a = 9b ⇒ 2:9

2) a + b = 3b ⇒ a = 2b ⇒ 1:2

Answers: 2:9 and 1:2

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3 years ago