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Answer:
∠CAB = 28°
∠DAC = 64°
Step-by-step explanation:
What you do in each case is make use of the relationships you know about angles in a triangle and around parallel lines. You can also use the relationships you know about diagonals in a rectangle, and the triangles they create.
<u>Left</u>
Take advantage of the fact that ∆AEB is isosceles, so the angles at A and B in that triangle are the same. If we call that angle measure x, then we have the sum of angles in that triangle is ...
x + x + ∠AEB = 180°
2x = 180° -124° = 56°
x = 28°
The measure of angle CAB is 28°.
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<u>Right</u>
Sides AD and BC are parallel, so diagonal AC can be considered a transversal. The two angles we're concerned with are alternate interior angles, so are congruent.
∠BCA = ∠DAC = 64°
The measure of angle DAC is 64°.
(Another way to look at this is that triangles BCE and DAE are congruent isosceles triangles, so corresponding angles are congruent.)
When a linear equation is in the form y = mx + c, the c, or constant, is the intercept on the y axis, meaning it crosses the y axis at (0, 1).
The gradient (1/3 in this case) is how much the y increments (or decrements) per increase of 1 of the value of x.
This would mean that there would be one point at (0, 1), and another at (3, 2). Draw a line from these two points and beyond, and that is the graph sketched.
Answer:
![y=\frac{c}{\sqrt[]{x^2+1} }](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bc%7D%7B%5Csqrt%5B%5D%7Bx%5E2%2B1%7D%20%7D)
Step-by-step explanation:
(1 + x²)dy +xydx= 0
Integrate both side
![lny=-\frac{1}{2} ln(x^2+1)+c\\y=\frac{c}{\sqrt[]{x^2+1} }](https://tex.z-dn.net/?f=lny%3D-%5Cfrac%7B1%7D%7B2%7D%20ln%28x%5E2%2B1%29%2Bc%5C%5Cy%3D%5Cfrac%7Bc%7D%7B%5Csqrt%5B%5D%7Bx%5E2%2B1%7D%20%7D)
