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erik [133]
3 years ago
11

Given two angles that measure 50 degrees and 80 degrees and side that measures 4 feet, how many triangles, if any, can be constr

ucted?
Mathematics
1 answer:
kirza4 [7]3 years ago
3 0
The angles are the only constraint here that counts.  If one of the three interior angles of a supposed triangle is 50 degrees and another is 80 degrees, then the third angle must be 50 degrees.  Thus, we have a 50-50-80 triangle, which is isosceles though not a right triangle.  If 4 feet is a measure of one of the equal sides of a supposed triangle, then obviously the adjacent side also has measure 4 ft.

The set of angles remains the same (50-50-80), but subject to the constraint mentioned above, the measure of any one of the sides has infinitely many possible values, so long as those values are positive.

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im struggling alot on my homework page and here is one of the questions i need help on can someone please help!?
zhuklara [117]

Answer:

6

Step-by-step explanation:

Express the equation in slope- intercept form

y = mx + b ( where m is the slope and b the y- intercept )

Here m = - 0.5, thus

y = - 0.5x + b

To find b substitute (10, 1) into the equation

1 = - 5 + b ⇒ b = 1 + 5 = 6 ← y- intercept

7 0
3 years ago
Using Laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.
professor190 [17]

Answer:

The solution to the differential equation is

X(s)=\cfrac 1{6}  -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular  solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

x''(t)+4x'(t)+6x(t)=1-e^t

So we will get

s^2 X(s)-sx(0)-x'(0)+4(sX(s)-x(0))+6X(s)=\cfrac 1s -\cfrac1{s-1}

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}

Simplifying

s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}

s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}

Moving all terms that do not have X(s) to the other side

s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5

Factoring X(s) and moving the rest to the other side.

X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5

X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}

So if we multiply both sides by the entire denominator we get

1=A(s^2+4s+6) +  (Bs+C)s

At this point we can find the value of A fast if we plug s = 0, so we get

1=A(6)+0

So the value of A is

A = \cfrac 16

We can replace that on the previous equation and multiply all terms by 6

1=\cfrac 16(s^2+4s+6) +  (Bs+C)s

6=s^2+4s+6 +  6Bs^2+6Cs

We can simplify a bit

-s^2-4s=  6Bs^2+6Cs

And by comparing coefficients we can tell the values of B and C

-1= 6B\\B=-1/6\\-4=6C\\C=-4/6

So the separated fraction will be

\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}

We can repeat the process for the second fraction.

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}

Multiplying by the entire denominator give us

1=A(s^2+4s+6) + (Bs+C)(s-1)

We can plug the value of s = 1 to find A fast.

1=A(11) + 0

So we get

A = \cfrac1{11}

We can replace that on the previous equation and multiply all terms by 11

1=\cfrac 1{11}(s^2+4s+6) + (Bs+C)(s-1)

11=s^2+4s+6 + 11Bs^2+11Cs-11Bs-11C

Simplifying

-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C

And by comparing coefficients we can tell the values of B and C.

-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C\\-1=11B\\B=-\cfrac{1}{11}\\5=-11C\\C=-\cfrac{5}{11}

So the separated fraction will be

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}

So far replacing both expanded fractions on the solution

X(s)=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6} -\cfrac {1/11}{s-1} -\cfrac {-s/11-5/11}{s^2+4s+6}+\cfrac {s+5}{s^2 +4s+6}

We can combine the fractions with the same denominator

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}

Simplifying give us

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{s^2 +4s+6}

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

s^2+4s+6 = s^2 +4s+4-4+6

We are adding half of the middle term but squared, so the first 3 terms become the perfect  square, that is

=(s+2)^2+2

So we get

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{(s+2)^2 +(\sqrt 2)^2}

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2-2)/66+316 /66}{(s+2)^2 +(\sqrt 2)^2}\\X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66+194 /66}{(s+2)^2 +(\sqrt 2)^2}

Lastly we can split the fraction one more

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66}{(s+2)^2 +(\sqrt 2)^2}+\cfrac {194 /66}{(s+2)^2 +(\sqrt 2)^2}

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

\boxed{X(s)=\cfrac 1{6}  -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)}

6 0
3 years ago
Write the vaule of the digit 9
snow_lady [41]
It's in the ones place
4 0
3 years ago
What does "place value" mean? Give specific examples of systems that use place value and those that do not.
zaharov [31]

Answer:

Place value can be defined as the value represented by a digit in a number on the basis of its position in the number.

Step-by-step explanation:

In the number 456, the "5" has a face value of 5. Value = The place value of the digit times its face value. Example: Find the value of the digits 6, 8 and 4 in the number 684.

6 0
3 years ago
I know this is silly but I wasn’t fought correctly how to read clocks and I’m struggling right now.
Gnoma [55]

Answer:

its 10:25 :)

Step-by-step explanation:

hope this helps!

4 0
3 years ago
Read 2 more answers
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