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3 years ago

Help Pls if you want!!!!

Mathematics

2 answers:

Tom [10]3 years ago

7 0

Answer:

Its 2.5 percent ladd enjoy

Step-by-step explanation:

Bezzdna [24]3 years ago

6 0

Answer:

2 per team and 2 left

Step-by-step explanation:

10 can't split into 4 evenly so, you can only give 2 to each team causing there to be 2 left.

4*2 = 8

10-8 = 2

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Find the area of the sector formed by central angle

Alex777 [14]

Answer: 0.2pi

Step-by-step explanation:

1. Find the area of the entire circle

2. Set up a proportion that compares the relationship of the Area of sector and the Area of circle to the Arc measure and the circle measure

3. Solve!

6 0

3 years ago

Please help me tysm!

Nikolay [14]

Answer:

Step-by-step explanation:

I wrote down the subject with the number of students. I looked one by one to see if the number of students with the subject matched with the graph. Then I finally found that the last graph matched with what I written down.

7 0

3 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

Let A=(1,2,3,4,5)which of the following sets are equal to A?

Rom4ik [11]

Answer:

1. (4,1,2,3,5)

Step-by-step explanation:

Since all the members of set A are present in "1"

5 0

3 years ago

Write the equation in slope-intercept form of the line that contains the points (4, -7) and (0, 5).

LenaWriter [7]

The slope intercept form is y=mx+b
The slope formula is m=(y2-y1)/(x2-x1)
So; 5-(-7) / 0-4= -3
Then you use one of the points to find b; I’ll use 0 and 5 (the first number is x and the second number is y)
5=(-3)(0)+b
5=0+b
5=b
Finally you plug in your two values
Y= -3x+5

Hope this helps :)

4 0

3 years ago

Read 2 more answers