Answer:
23.5(to the nearest tenth)
Step-by-step explanation:
pythagoras theorem states
c^2= a^2+ b^2
a= 23
b=5
c^2= 529+25
c^2= 554
c= square root of 554
c = 23.5 (approximately)
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Responder:
La temperatura bajó 12 °
Explicación paso a paso:
Dado que:
Temperatura en la noche = + 2 ° (2
Temperatura de la mañana = - 12 (12 ° bajo cero).
La diferencia de temperatura entre la noche y la madrugada:
Temperatura nocturna - temperatura por la mañana temprano
2 - (-12)
2 + 12 = 14
Hay una diferencia de temperatura de 12 °. La temperatura bajó 12 ° entre la noche y la madrugada.
What we know is that the sum of a triangle's internal angles are 180º, and a straight angle makes 180º.
first, we have to find the third internal angle(let's call it n) so that we can use it to find the exterior angle.
(3x + 20) + (4x + 5) + n = 180
simplify:
7x + 25 + n = 180
7x + n = 155
n = 155 - 7x
Now we can add it to the external angle to find x.
(155 - 7x) + (8x + 15) = 180
simplify:
170 + 8x - 7x = 180
170 + x = 180
x = 10
Now we can substitute it to the external angle.
8 x 10 + 15 = 95
the exterior angle is 95º.
<span>For this case we have the following equation:</span>
<span> (x + 2) 2 + y2 = 10</span>
<span> We must remember that the standard equation of the circle is given by:</span>
<span> (x - h) 2 + (y - k) 2 = r2</span>
<span> Where r is the radius.</span>
<span> Therefore, in the given equation the radius is:</span>
<span> r ^ 2 = 10</span>
<span> Clearing we have:</span>
<span> R = root (10)</span>
<span> Answer:</span>
<span> The length of the radius of the circle is:</span>
<span><span> √ (10)</span></span>
Answer: 41.1
Step-by-step explanation:
To find the value of x, we need to use SOH-CAH-TOA. SOH-CAH-TOA is since, cosine, and tangent. The O, A, H stands for opposite, adjacent, and hypotenuse respectively.
Looking at the figure, we see 47°. The labelled angles are adjacent and the hypotenuse. Therefore, we use CAH or cosine.
[multiply both sides by x]
[divide both sides by cos(47)]
When we plug that into a calculator, we get x=41.1.
