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3 years ago

Look at the figure. Which pair of points is coplanar with points V and W?

Mathematics

2 answers:

Ann [662]3 years ago

6 0

Answer:

The pair of points which are coplanar with points V and W are:

R and S

Step-by-step explanation:

Two or more points are said to be coplanar if they all lie in the same plane.

Here by looking at the figure we observe that the points V and W along with R and S would lie in the plane labeled as VWSR.

( Also V and W are coplanar with U and X since we get a plane UVWX such that all the four points lie on that plane)

Hence, the answer is:

R and S

aleksley [76]3 years ago

4 0

The answer is r and s because it’s in the same plane.

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erma4kov [3.2K]

Answer:

the answer is 262 days

Step-by-step explanation:

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2 years ago

What is the value ofcos pie value

Hitman42 [59]

Cos(pi) = -1

You should have this memorized along with other values such as
sin(0) = 0
cos(0) = 1
sin(pi/2) = 1
cos(pi/2) = 0
sin(pi) = 0
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cos(3pi/2) = 0

and remember that tan = sin/cos

3 0

2 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Mr. Kramer's patio is in the shape of a trapezoid. The trapezoid and its dimensions are shown below. What is the area of the pat

BigorU [14]

Answer:

C 315

Step-by-step explanation:

(12 + 30) * 15/2 = 315

5 0

2 years ago

I am a 2 digit number the sum of my digits is 12. the ones digit is twice the tens digit? who am i?

Bas_tet [7]

Answer:

Step-by-step explanation:

let a be the 1's digit and b the 10's digit, then

a = 2b ← twice the 10's digit, thus

sum = 2b + b = 12, that is

3b = 12 ( divide both sides by 3 )

b = 4

and a = 2b = 2 × 4 = 8

The 2 digit number is 84

5 0

3 years ago