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2 years ago

The sin (0) = - and lies in Quadrant III. Find the exact values of the sine

Mathematics

1 answer:

CaHeK987 [17]2 years ago

5 0

Answer:

Can you give exanation

I don't understand

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a branch of mathematics that deals with the study of collecting, analyzing, interpreting, presenting, and organizing data in a p

saw5 [17]

Answer:

Statistics

Step-by-step explanation:

is the application of probability theory, a branch of mathematics, to statistics, as opposed to techniques for collecting statistical data.
the study and manipulation of data, including ways to gather, review, analyze, and draw conclusions from data.

7 0

2 years ago

Read 2 more answers

clark made a model of his house. his house is 30 1/2 feet long. the dimensions of the model were 1/25 the dimensions of clarks a

balandron [24]

Answer:

1.22 feet, or 1 11/50 ft

Step-by-step explanation:

Find 1/25 th of 30 1/2 feet:

1 61 ft

----- * ---------- = 1.22 feet, or 1 11/50 ft

25 2

The length of the model is 1.22 feet, or 1 11/50 ft

7 0

3 years ago

Darrius bought 3 gallons of ice cream for $9.50 how much will it cost him if he wants 7 gallons?

Charra [1.4K]

Answer:

Its going to cost him 22.167

Step-by-step explanation:

Divide $9.50 by 3 = 3.167

multiply it by 7= 22.167

3 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

Given An = 5a(n-1) where a1= 2. What are the first 4 terms

nexus9112 [7]

Answer:

2, 10, 50, 250

Step-by-step explanation:

Using the formula with a₁ = 2 , then

a₂ = 5a₁ = 5 × 2 = 10

a₃ = 5a₂ = 5 × 10 = 50

a₄ = 5a₃ = 5 × 50 = 250

The first 4 terms are 2, 10, 50, 250

8 0

3 years ago