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BabaBlast [244]

3 years ago

12

Two endpoints of the diagonal of a parallelogram are k(0,3) and l(4,1). what is the length of the diagonal

1 answer:

posledela3 years ago

7 0

Distance = sqrt (x2 - x1)^2 + (y2 - y1)^2
(0,3).....x1 = 0 and y1 = 3
(4,1)....x2 = 4 and y2 = 1
now we sub and solve
d = sqrt (4 - 0)^2 + 1 - 3)^2
d = sqrt (4^2) + (-2^2)
d = sqrt (16 + 4)
d = sqrt 20
d = 4.472 <==

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Wewaii [24]

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If there are a total of 5 pieces and you get 3 pieces of the pie, is that larger than getting 1 piece of the pie? Yes!

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If i have a stadium and it holds 128 people on the second floor, and 5874 on the first floor how many people can I have at the s

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3 years ago

Use the spinner to find the theoretical probability of the event. Write your answer as a fraction or a percent rounded to the ne

Rama09 [41]

Complete question :

Picture relating to the question is attached below.

Answer:

1 / 3

Step-by-step explanation:

The Theoretical probability of an event is defined as :

P(event) = required outcome / Total possible outcomes

Based on the picture of the attached spinner below :

P(red) :

Required outcome = number of red partitions = 2

Total possible outcomes = total Number of partitions = 6

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3 0

3 years ago

What is the value of 3x^2 + 4y^2 if x = 2 y = 1

Ahat [919]

Answer:

16 is answer

Step-by-step explanation:

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6 0

3 years ago

Read 2 more answers

Sand is leaking through a small hole at the bottom of a conical funnel at the rate of 12cm3/s . if the radius of the funnel is 8

Bezzdna [24]

Answer:

h ≅ 16.29 cm

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$

Step-by-step explanation:

From the conical funnel;

Let the adjacent line at the base of the cone be radius (r) and the opposite ( vertical length) be the height (h)

Then;

$tan \theta= \dfrac{r}{h}$

$tan \theta= \dfrac{8}{16}$

$tan \theta= \dfrac{1}{2}$

∴

$\dfrac{r}{h} = \dfrac{1}{2}$

$r = \dfrac{h}{2}$

The volume (V) of the cone is expressed as:

$V = \dfrac{1}{3}\pi r ^2 h$

$V = \dfrac{1}{3}\pi (\dfrac{h}{2})^2 h$

$V = \dfrac{\pi h^3}{3\times 4}$

$\dfrac{dv}{dt} = \dfrac{3 \pi h^2}{3 \times 4} \dfrac{dh}{dt}$

$\dfrac{dv}{dt} = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

Given that:

$\dfrac{dv}{dt} = 12 \pi$

Then:

$-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$- \int 48 \ dt = \int h^2 \ dh$

$\dfrac{h^3}{3}= -48 t + c$

$At \ t = 0 \ ; h = 0$

∴

$\dfrac{0^3}{3} = -48(0) + c$

c = 0

So;

$\dfrac{h^3}{3}= -48 t + 0$

$\dfrac{h^3}{3}= -48 t$

t = 30

$\implies h = (-48(30))^{1/3}$

h = 16.2865

h ≅ 16.29 cm

Thus;

from $-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$\dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} } =\dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} }$

$\dfrac{dh}{dt}=\dfrac{-12 \times 4} { {16.29^2} }$

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$

7 0

3 years ago