There are 8 students in each class and there are 5 classes
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I hope the choices for the numerators of the solutions are given.
I am showing the complete work to find the solutions of this equation , it will help you to find an answer of your question based on this solution.
The standard form of a quadratic equation is :
ax² + bx + c = 0
And the quadratic formula is:
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
So, first step is to compare the given equation with the above equation to get the value of a, b and c.
So, a = 10, b = -19 and c = 6.
Next step is to plug in these values in the above formula. Therefore,
So,
So,
Hope this helps you!
Answer:
see the attachments for the two solutions
Step-by-step explanation:
When the given angle is opposite the shorter of the given sides, there will generally be two solutions. The exception is the case where the triangle is a right triangle (the ratio of the given sides is equal to the sine of the given angle). If the given angle is opposite the longer of the given sides, there is only one solution.
When a side and its opposite angle are given, as here, the law of sines can be used to solve the triangle(s). When the given angle is included between two given sides, the law of cosines can be used to solve the (one) triangle.
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Here, the law of sines can be used to solve the triangle:
A = arcsin(a/c·sin(C)) = arcsin(25/24·sin(70°)) = 78.19° or 101.81°
B = 180° -70° -A = 31.81° or 8.19°
b = 24·sin(B)/sin(70°) = 13.46 or 3.64
Answer:
if u ll v then this angel is the twin of the one below and is 180-72
Step-by-step explanation: