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Anna11 [10]
4 years ago
7

I need help on this problem

Mathematics
2 answers:
chubhunter [2.5K]4 years ago
6 0
The value of x
3x°+46°+(2x-1°)=180°(angle sum of △)
5x+45=180°
5x=180°-45
x=135÷5
x=27
x is 27

put x into angle B,we have
2(27)-1
=53

thus angleB is 53°

hope it helps c:
likoan [24]4 years ago
3 0
In Euclidean geometry (the geometry you're probably familiar with), the interior angles of a triangle always add to 180°.

With that knowledge, we can set up an equation that lets us solve for x based on our given angles by setting their sum equal to 180.

3x+(2x-1)+46=180\\5x+45=180\\5x=135\\x=27

To find ∠B, we simply substitute the x for 27:

2x-1=2(27)-1=54-1=53

So, m∠B = 53°.
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it took Jenny and her family one hour to drive to her grandmother's house her grandmother lives about 45 miles away at what rate
gayaneshka [121]

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The family traveled at 45 miles per hour.

Step-by-step explanation:

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3 years ago
An equation of a hyperbola is given.
siniylev [52]

Answer:

a)

The vertices are \left(3,\:0\right),\:\left(-3,\:0\right).

The foci are \left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right).

The asymptotes are y=2x,\:y=-2x.

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1 is the standard equation for a right-left facing hyperbola with center \left(h,\:k\right).

a)

The vertices\:\left(h+a,\:k\right),\:\left(h-a,\:k\right) are the two bending points of the hyperbola with center \:\left(h,\:k\right) and semi-axis a, b.

Therefore,

\frac{x^2}{9}-\frac{y^2}{36}=1, is a right-left Hyperbola with \:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6 and vertices \left(3,\:0\right),\:\left(-3,\:0\right).

For a right-left facing hyperbola, the Foci (focus points) are defined as \left(h+c,\:k\right),\:\left(h-c,\:k\right) where c=\sqrt{a^2+b^2} is the distance from the center \left(h,\:k\right) to a focus.

Therefore,

\frac{x^2}{9}-\frac{y^2}{36}=1, is a right-left Hyperbola with \:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6 c=\sqrt{3^2+6^2}= 3\sqrt{5} and foci \left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)

The asymptotes are the lines the hyperbola tends to at \pm \infty. For right-left hyperbola the asymptotes are: y=\pm \frac{b}{a}\left(x-h\right)+k

Therefore,

\frac{x^2}{9}-\frac{y^2}{36}=1, is a right-left Hyperbola with \:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6 and asymptotes

y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x

b) The length of the transverse axis is given by 2a. Therefore, the lenght is 6.

c) See below.

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27 = 9^(3/2) 
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so x = 4/3
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4 years ago
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