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mezya [45]
3 years ago
12

If the sum of the measure of two angles in a triangle is 101, then the measure

Mathematics
2 answers:
nalin [4]3 years ago
8 0
The third angle is 79
180-101=79
LiRa [457]3 years ago
5 0

Answer:

79 degrees.

Step-by-step explanation:

It is just a rule of trigonometry that all the angles inside ANY triangle will add up to 180 degrees.

There's only 3 possible angles in a triangle, so if you know what the sum of 2 are, its easy to find the last one.

All you have to do is 180 - 101, which equals 79 degrees.

Hope this helped : )

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The graph shows the distance traveled by two runners over several minutes,
ElenaW [278]

Runner B is going slower.

7 0
3 years ago
Read 2 more answers
How many solutions does this linear system have?
ziro4ka [17]

Answer: C is correct, No solution

Step-by-step explanation:

Let’s simplify the second equation,

8x-4y=-20

We can divide the whole equation by 4...

8x/4-4y/4=-20/4

Which becomes....

2x-y=-5

Now let’s move things around to see if it’s the same equation as the one on top...

-y=-2x-5

y=2x+5

Now we need to solve the systems...

y=2x-5

y=2x+5

Since their slopes are both 2 and their y-intercepts are not identical, the answer is no solution. The two lines will continue on without ever crossing because they have the same slope.

4 0
3 years ago
The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$, then what are the $x$-intercepts o
Alex_Xolod [135]

Your calculator's cubic regression function can tell you the equation is

... f(x) = 2x³ + 5x² -12x = x(x +4)(2x-3)

The x-intercepts are -4, 0, +1.5.

_____

If you want to solve this "by hand", you can first of all recognize that since there is an x-intercept at 0, the cubic will only have three coefficients. That is, you can write the equation as

... f(x) = ax³ + bx² + cx

Substituting the given points (except (0, 0)) gives three linear equations in a, b, c.

... -a +b -c = 15 . . . . . for x=-1

... a + b + c = -5 . . . . for x=1

... 8a +4b +2c = 12 . . for x=2

adding the first two equations gives 2b=10, or b=5. Now, you can reduce the system to

... a + c = -10

... 4a +c = -4

Subtracting the first of these equations from the second gives 3a=6, or a=2. That tells you c=-12 (from a+c=10).

Then your equation is

... f(x) = x(2x² +5x -12)

Factoring by any of the usual techniques, or graphing, or using the quadratic formula will tell you the zeros (x-intercepts) are as above.

_____

Since the input values are sequential, you can also develop the function from differences of the output values. Those are 15, 0, -5, 12. First differences are -15, -5, +17. Second differences are +10, +22. The third difference is 12. Using the first of these differences in appropriate places in the interpolating polynomial formula, we have

... f(x) = 15 + (x+1)(-15 + (x)/2·(10 + (x-1)/3·(12))) = 2x³ +5x² -12x . . . . as above

4 0
2 years ago
Which expression makes the equation true?
valina [46]
Equation is y^20 = (   )^2

Answer: 

y to the power of 10. 
4 0
3 years ago
Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2
iren2701 [21]
Hello there, hope I can help!

I assume you mean L\left\{\frac{ekt+e-kt}{2}\right\}
With that, let's begin

\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \  L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}

\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}
\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b
L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}
\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}

L\left\{t\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \  L\left\{t\right\}=\frac{1}{s^2}

L\left\{\frac{e}{2}\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \  L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \  \frac{e}{2s}

\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}

\frac{ek}{2}\cdot \frac{1}{s^2}  \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{ek\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{ek}{2s^2}

\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{k\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{k}{2s^2}

\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}

Hope this helps!
3 0
3 years ago
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