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4 years ago

PLZ HURRY! Identify a Venn diagram to show how many students in the sixth grade belong to both cultural and sports team.

Mathematics

1 answer:

REY [17]4 years ago

5 0

On one side of the venn diagram label it cultural team on the other side label sports team and in the middle put both

cultural will have: 5 students for YES (idk if you need for NO but if so, it’s 6)

sports will have: 5 students for YES (6 for NO)

both will have: 2 SETS of students for YES which is 4 students (3 SETS of students for NO which is 6 students)

i hope it’s correct <3

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Given the data below:

Leno4ka [110]

Answer:

5.16

Step-by-step explanation:

Since after x=3, value of y starts decreasing for increasing values of x, let's choose. Also we need to find f(x) for x=3.4

x=3,4,5 and f(x)=7,3,1 to find lagrange polynomial

P(x)= ((x-x2)(x-x3)y1)/((x1-x2)(x1-x3) + ((x-x1)(x-x3)y2)/((x2-x1)(x2-x3) + ((x-x1)(x-x2)y3)/((x3-x1)(x3-x2)

P(x)= ((x-4)(x-5)7)/((3-4)(3-5)) + ((x-3)(x-5)3)/((4-3)(4-5)) + ((x-3)(x-4)1)/((5-3)(5-4))

P(x)= x² -11x+31

P(3.4)= 5.16

7 0

3 years ago

Help me out!!

vekshin1

Answer:

~Exponential decay because the base is less than 1~

Step-by-step explanation:

I just took the test and got a 100% :) hope this helped! Good luck!

8 0

3 years ago

Find the area of ABC with the side lengths a=8 and b= 12, and the included angle C =84 degrees. Round your answer to the nearest

spin [16.1K]

Answer:

47.7 square units

Step-by-step explanation:

The area of triangle ABC with two given sides and an inclusive angle =

1/2absinC

Where : a = 8, b = 12, C = 84

Area of Triangle ABC = 1/2 × 8 × 12 × sin 84

Area of Triangle ABC = 47.737050978 square units

Approximately to the nearest tenth = 47.7 square units.

6 0

3 years ago

Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)

vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

$y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\$

Apply substitution to get

$y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right) + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x) + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x) + \left(a - 6b\right)\cos(x) = \sin(x)\\\\$

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0

3 years ago

g Given the system of equations: 15 c1 – 5 c2 – c3 = 2400 - 5 c1 + 18 c2 – 6 c3 = 3500 - 4 c1 - c2 + 12 c3 = 4000 (a) Calculate

denpristay [2]

Answer

(a)

$A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}$

(b)

$A^{-1} b = \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}$

Step-by-step explanation:

Remember that when you want to solve a problem like this, you express the equation as following

$Ax = b$

So, if you know the inverse of $A$ then

$x = A^{-1} b$

For this case

$A = \begin{pmatrix} 15 && 5 && -1 \\ -5 && 18 && -6 \\ -4 && -1 && 12 \end{pmatrix}$

Now for this case the inverse of A would be

$A^{-1} = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}$

Then when you multiply with the vector solution

$A^{-1} b = \frac{1}{3493} \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix} * \begin{pmatrix} 2400 \\ 3500 \\ 4000 \end{pmatrix} = \frac{1}{3493} \begin{pmatrix} 249500 \\ 1197600 \\ 1347300 \end{pmatrix}\\\\\\= \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}$

So from that information you can conclude that the solution to the system of equations is x = 500/7 y = 2400/7 and z = 2700/7

7 0

3 years ago