Answer:
isosceles triangle
Step-by-step explanation:
An isosceles triangle therefore has both two equal sides and two equal angles. The name derives from the Greek iso (same) and skelos (leg). A triangle with all sides equal is called an equilateral triangle, and a triangle with no sides equal is called a scalene triangle.
The literal equation for x is g/4+5y
<h3>Subject of formula</h3>
This is a way of representing a variable in terms of another
Given the equation below;
g=4x+5xy
Factor out x to have
g=x(4 + 5y)
Divide both sides by 4 +5y to have:
g/4+5y = x
x = g/4+5y
Hence the literal equation for x is g/4+5y
Learn more on subject of formula here: brainly.com/question/21140562
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Answer:
The measures of angles of triangle MNP are
Step-by-step explanation:
<u><em>The picture of the question in the attached figure</em></u>
step 1
Find the measure of arcs AB, BC and AC
we know that
The inscribed angle is half that of the arc it comprises.
so
step 2
Find the measure of angle M
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
![M=\frac{1}{2}[arc\ AB+arc\ BC-arc\ AC]](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AB%2Barc%5C%20BC-arc%5C%20AC%5D)
substitute
![M=\frac{1}{2}[2\gamma+2\alpha-2\beta]\\M=[\gamma+\alpha-\beta]](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1%7D%7B2%7D%5B2%5Cgamma%2B2%5Calpha-2%5Cbeta%5D%5C%5CM%3D%5B%5Cgamma%2B%5Calpha-%5Cbeta%5D)
step 3
Find the measure of angle N
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
![N=\frac{1}{2}[arc\ AC+arc\ BC-arc\ AB]](https://tex.z-dn.net/?f=N%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AC%2Barc%5C%20BC-arc%5C%20AB%5D)
substitute
![N=\frac{1}{2}[2\beta+2\alpha-2\gamma]\\N=[\beta+\alpha-\gamma]](https://tex.z-dn.net/?f=N%3D%5Cfrac%7B1%7D%7B2%7D%5B2%5Cbeta%2B2%5Calpha-2%5Cgamma%5D%5C%5CN%3D%5B%5Cbeta%2B%5Calpha-%5Cgamma%5D)
step 4
Find the measure of angle P
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
![P=\frac{1}{2}[arc\ AC+arc\ AB-arc\ BC]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AC%2Barc%5C%20AB-arc%5C%20BC%5D)
substitute
![P=\frac{1}{2}[2\beta+2\gamma-2\alpha]\\P=[\beta+\gamma-\alpha]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B1%7D%7B2%7D%5B2%5Cbeta%2B2%5Cgamma-2%5Calpha%5D%5C%5CP%3D%5B%5Cbeta%2B%5Cgamma-%5Calpha%5D)
I think it’s not equal because if you took (1)(2)(3). You would need to multiply them not add them. It would be 9 = (1,2,3)
Try this solution:dy/dx=y⁴cosx; y⁻⁴dy=cosxdx;
