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SpyIntel [72]
3 years ago
10

Brainly for correct answer!

Mathematics
2 answers:
DochEvi [55]3 years ago
8 0

Answer:

It is b

Step-by-step explanation:

Like I said in the previous answer the direction of the "<" sign can tell you, and since it is the " < or equal to " sign it is a solid line not dotted. So therefore its b.

Cerrena [4.2K]3 years ago
6 0

Answer:

I think B sorry if wrong

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Find the sum. Write your answer in simplest form.
gulaghasi [49]

Answer:

the answer for your question is 9 31/42

3 0
3 years ago
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Please answer this ASAP
Amanda [17]

Answer: 2x and y= 4+

Step-by-step explanation:

x = 2 times 2 and so on....

y = 4 plus 4 plus 4 etc....

3 0
3 years ago
A surveyor, Toby, measures the distance between two landmarks and the point where he stands. He also measured the angles between
vagabundo [1.1K]
What we know so far:
Side 1 = 55m
Side 2 = 65m
Angle 1 = 40°
Angle 2 = 30°

What we are looking for:
Toby's Angle = ?
The distance x = ?

We need to look for Toby's angle so that we can solve for the distance x by assuming that the whole figure is a SAS (Side Angle Side) triangle.

Solving for Toby's Angle:
We know for a fact that the sum of all the angles of a triangle is 180°; therefore,
180° - (Side 1 + Side 2) = Toby's Angle
Toby's Angle  = 180° - (40° + 30°)
Toby's Angle = 110°

Since we already have Toby's angle, we can now solve for the distance x by using the law of cosines r² = p²+ q²<span>− 2pq cos R where r is x, p is Side1, q is Side2, and R is Toby's Angle.
</span>
x² = Side1² + Side2² - 2[(Side1)(Side2)] cos(Toby's Angle)
x² = 55² + 65² - 2[(55)(65)] cos(110°)
x² = 3025 + 4225 -7150[cos(110°)]
x² = 7250 - 2445.44
x = √4804.56
x = 69.31m

∴The distance, x, between two landmarks is 69.31m
8 0
3 years ago
   2. By selling an article for ₹1636.25, a dealer gains ₹96.25. Find his gain percentage.  ​
madreJ [45]

Answer is..

5.88%

(96.25/1636.25)%

= 5.88%

4 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 <em>c)</em>

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

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

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 <em>A</em> and <em>B</em> 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
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