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3 years ago

Someone please help asap!!! How do I know where to shade at?

Mathematics

1 answer:

alexira [117]3 years ago

6 0

Shade the top side of the boundary line if you have the inequality symbols > or ≥. Shade the bottom side of the boundary line if you have the inequality symbols < or ≤.

$\large\bold\red{Answered By Ayush8928 }$

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Here is a central angle that measures 1.5 radians. Select all true statements. Question 1 options: A) The area of the whole circ

KATRIN_1 [288]

Answer:

D, E

Step-by-step explanation:

The applicable formula is ...

s = rθ

so, for θ = 1.5, the arc length is 1.5 times the radius. The true statements are ...

D) The length of the arc defined by the angle is 1.5 times longer than the radius.

E) The ratio of arc length to radius is 1.5.

8 0

3 years ago

Solve. Find all solutions in [0,2).5 secx cotx+5 secx+ cotx+1=0

Dmitrij [34]

$\begin{gathered} 5sec(x)cot(x)+5sec(x)+cot(x)+1=0 \\ Factoring \\ (cot(x)+1)(5sec(x)+1)=0 \\ cot(x)+1=0 \\ cot(x)=-1 \\ x=cot^{-1}(1) \\ x=\frac{3\pi}{4} \\ \\ 5sec(x)+1=0 \\ 5sec(x)=-1 \\ sec(x)=\frac{-1}{5},\text{ there is no solution} \\ Hence \\ All\text{ solution are }\frac{3\pi}{4}+\pi n,\text{ where n=1,2,3...} \\ \\ \end{gathered}$

5 0

1 year 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

HELP ME SOMEBODY PLSS

OverLord2011 [107]

Answer:

-6/8

Step-by-step explanation:

The rate of change is the same as the slope (I don't know if you have learned that term yet), but basically its the change in y's over the change in x's. to find this out, all you have to do is count from the point on the graph that is the farthest right (in this case, (6,1)), up to the y coordinate of the point farthest to the left (-2, 7) which is 6, and then count the number of spaces between the x coordinates, which is 8. put the rise (up and down count) over the run (horizontal count) in a fraction form, which gives you 6/8, and then notice how the line is pointing towards the bottom RIGHT hand corner making the line negative, so you need to make the rate of change negative, giving you the final answer of -6/8.

4 0

3 years ago

How do you solve 3/5 times 2/3

UkoKoshka [18]

Hello! And thank you for your question!

First use the rule: A/b x c/d = ac/bd

3 x 2 over 5 x 3

Simplify 3 x 2:

6 over 5 x 3

Simplify 5 x 3:

6/15

Simplify further:

2/5

Final Answer:

2/5

3 0

3 years ago