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3 years ago

12

This should be easy for me but i forgot how

1 answer:

astraxan [27]3 years ago

5 0

E is currently at (-5, 1). However, the dilation multiplies the distance between this point and the origin by six, resulting in (-30, 6) as the final coordinates.

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Guys look at the pic of the park and read the question I need this ASAP WITH ALL STEPS the choices are

Alik [6]

Answer:

A. 821.5 Square Yards

Step-by-step explanation:

30 x 30= 900

900 is the area for the whole park

To find the pool area you need to do pi times Radius squared

so 3.14 x 5 x 5

That equals 78.5 for the pool area.

So to find the grass area you need to subtract the pool area from the park

900 - 78.5 = 821.5 square yards!

5 0

3 years ago

2x-5y+5z=-10<br> 5x-4y+3z=-19<br> X-y+5z=17

yarga [219]

Answer:

x = -125/71 , y = 448/71 , z = 356/71

Step-by-step explanation:

Solve the following system:

{2 x - 5 y + 5 z = -10 | (equation 1)

5 x - 4 y + 3 z = -19 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Swap equation 1 with equation 2:

{5 x - 4 y + 3 z = -19 | (equation 1)

2 x - 5 y + 5 z = -10 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Subtract 2/5 × (equation 1) from equation 2:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - (17 y)/5 + (19 z)/5 = (-12)/5 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Multiply equation 2 by 5:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

x - y + 5 z = 17 | (equation 3)

Subtract 1/5 × (equation 1) from equation 3:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x - y/5 + (22 z)/5 = 104/5 | (equation 3)

Multiply equation 3 by 5:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x - y + 22 z = 104 | (equation 3)

Subtract 1/17 × (equation 2) from equation 3:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x+0 y+(355 z)/17 = 1780/17 | (equation 3)

Multiply equation 3 by 17/5:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x+0 y+71 z = 356 | (equation 3)

Divide equation 3 by 71:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y + 19 z = -12 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Subtract 19 × (equation 3) from equation 2:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x - 17 y+0 z = (-7616)/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Divide equation 2 by -17:

{5 x - 4 y + 3 z = -19 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Add 4 × (equation 2) to equation 1:

{5 x + 0 y+3 z = 443/71 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{5 x+0 y+0 z = (-625)/71 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Divide equation 1 by 5:

{x+0 y+0 z = (-125)/71 | (equation 1)

0 x+y+0 z = 448/71 | (equation 2)

0 x+0 y+z = 356/71 | (equation 3)

Collect results:

Answer: {x = -125/71 , y = 448/71 , z = 356/71

5 0

3 years ago

A concession stand at an athletic event is trying to determine how much to sell cola and iced tea for in order to maximize reven

Cerrena [4.2K]

Solution :

Demand for cola : 100 – 34x + 5y

Demand for cola : 50 + 3x – 16y

Therefore, total revenue :

x(100 – 34x + 5y) + y(50 + 3x – 16y)

R(x,y) = $100x-34x^2+5xy+50y+3xy-16y^2$

$R(x,y) = 100x-34x^2+8xy+50y-16y^2$

In order to maximize the revenue, set

$R_x=0, \ \ \ R_y=0$

$R_x=\frac{dR }{dx} = 100-68x+8y$

$R_x=0$

$68x-8y=100$ .............(i)

$R_y=\frac{dR }{dx} = 50-32x+8y$

$R_y=0$

$8x-32y=-50$ .............(ii)

Solving (i) and (ii),

4 x (i) ⇒ 272x - 32y = 400

(ii) ⇒ (-<u>) 8x - 32y = -50 </u>

264x = 450

∴ $x=\frac{450}{264}=\frac{75}{44}$

$y=\frac{175}{88}$

So, x ≈ $ 1.70 and y = $ 1.99

R(1.70, 1.99) = $ 134.94

Thus, 1.70 dollars per cola

1.99 dollars per iced ted to maximize the revenue.

Maximum revenue = $ 134.94

4 0

3 years ago

What are the vertex focus and directrix of a parabola with equation x=y^2+14y-2

zimovet [89]

This is a sideways opening parabola, opening to the right to be more specific, since the leading coefficient is a positive 1. The rule for a focus and a directrix is that they are the same number of units from the vertex (in other words, the vertex is dead center between them), and that the vertex is on the same axis that the focus is. We need to find the vertex then to determine what the focus and the directrix are. We will complete the square on that to find the vertex. Begin by setting it equal to 0, then move the 2 over by addition to get $y^2+14y=2$ . Now we will complete the square on the y terms. Take half the linear term, square it, and add it to both sides. Our linear term is 14. Half of 14 is 7, and 7 squared is 49. So we add 49 to both sides. $y^2+14y+49=2+49$ , which of course simplifies to $y^2+14y+49=51$ . The purpose of this is to find the k coodinate of the vertex which will be revealed when we write the perfect square binomial we created during this process: $(y+7)^2=51$ . Moving the 51 back over by subtraction gives us $(y+7)^2-51=x$ . The vertex then is (-51,-7). The formula to find the focus using this vertex is $(h+ \frac{1}{4a},k)$ . As I stated quite a while ago, the leading coefficient on our parabola was a +1 so our "a" value is 1, and the focus is then found in $(-51+ \frac{1}{4},-7)$ which simplifies to $(-50.75, -7)$ . If the vertex is (-51, -7) and the focus is (-50.75, -7), then the distance between them is 1/4, or .25. That means that the directrix is also .25 units from the vertex, but in the other direction. Our directrix is a vertical line, and it will have the equaion x = -51.25. Summing up, your focus is (-50.75, -7) and your directrix is x = -51.25

7 0

4 years ago

The number of text messages sent daily by a student is a poisson random variable with parameter λ=5 .in a class with 20 independ

Rudik [331]

This problem is a combination of the Poisson distribution and binomial distribution.

First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961

For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
=6.18101*10^(-5) or approximately
=0.00006181

7 0

4 years ago