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Ymorist [56]
3 years ago
10

What is the value of y

Mathematics
2 answers:
Ilya [14]3 years ago
5 0

Answer: C

Step-by-step explanation: A straight angle is 180 degrees:

180-130 = 50

Triangle angles equal up to 180.

50 + 2y =180

y =65

nasty-shy [4]3 years ago
3 0
<h2>QUESTION -> What is the value of y</h2>

<h2>ANSWER :- </h2>

SUM OF INTERIOR OPPOSITE ANGLES = EXTERIOR ANGLE .

130 = y + y \\ 130 = 2y \\  \frac{130}{2}  = y \\  \frac{ \cancel{130}^ { \tiny  \:  \:  \:  \: 65} }{ \cancel{2}}  = y \\ y = 65   \degree

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Find the area of the parallelogram.
Zolol [24]

Answer:

14.4 km²

Step-by-step explanation:

A=b

h=4.8·3

≈14.4

3 0
2 years ago
where does the graph of the function y=tan(x) have asymptotes? at the values of x where cos(x)=0 at the values of x where sin(x)
Gnom [1K]
Answer: at the values where cos(x) = 0

Justification:

1) tan(x) = sin(x) / cos(x).

2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.

3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.

Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.

You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.



7 0
3 years ago
Read 2 more answers
Area of a triangle with points at (-9,5), (6,10), and (2,-10)
Ann [662]
First we are going to draw the triangle using the given coordinates. 
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

Distance from point A to point B:
d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2}
d_{AB}= \sqrt{(6+9)^2+(10-5)^2}
d_{AB}= \sqrt{(15)^2+(5)^2}
d_{AB}= \sqrt{225+25}
d_{AB}= \sqrt{250}
d_{AB}=15.81

Distance from point A to point C:
d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}
d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}
d_{AC}= \sqrt{11^2+(-15)^2}
d_{AC}= \sqrt{121+225}
d_{AC}= \sqrt{346}
d_{AC}= 18.60

Distance from point B from point C
d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}
d_{BC}= \sqrt{(-4)^2+(-20)^2}
d_{BC}= \sqrt{16+400}
d_{BC}= \sqrt{416}
d_{BC}=20.40

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
s= \frac{AB+AC+BC}{2}
s= \frac{15.81+18.60+20.40}{2}
s= \frac{54.81}{2}
s=27.41

Finally, to find the area of our triangle, we are going to use Heron's formula:
A= \sqrt{s(s-AB)(s-AC)(s-BC)}
A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)}
A= \sqrt{27.41(11.6)(8.81)(7.01)}
A=140.13

We can conclude that the perimeter of our triangle is 140.13 square units.

3 0
2 years ago
3x-2x+8=12; x=4 <br>Determine if the value of x is a solution to the equation<br>​
VARVARA [1.3K]

Step-by-step explanation:

3x - 2x + 8 = 12; x=4

Now, Evaluate the value of x

3(4) - 2(4) + 8 = 12

12 - 8 + 8 = 12

12 + 0 = 12

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4 0
2 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
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