These are the <span>xx</span> and <span>yy</span> intercepts of the equation <span><span>2x−5y=6</span><span>2x-5y=6</span></span>.x-intercept: <span><span>(3,0)</span><span>(3,0)</span></span>y-intercept: <span>(0,−<span>65</span><span>)
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Answer:
The perimeter is 
Step-by-step explanation:
we know that
A parallelogram is a quadrilateral where both pairs of opposite sides are parallel and equal
so
In this problem
PS=QR ----> equation A
SR=PQ ----> equation B
The perimeter of parallelogram PQRS is
P=PQ+QR+SR+PS ----> equation C
substitute equation A and equation B in equation C

we have


substitute in the formula of perimeter


Answer: The first equation is an equation of a parabola. The second equation is an equation of a line.
Explanation:
The first equation is,

In this equation the degree of y is 1 and the degree of x is 2. The degree of both variables are not same. Since the coefficients of y and higher degree of x is positive, therefore it is a graph of an upward parabola.
The second equation is,

In this equation the degree of x is 1 and the degree of y is 1. The degree of both variables are same. Since both variables have same degree which is 1, therefore it is linear equation and it forms a line.
Therefore, the first equation is an equation of a parabola. The second equation is an equation of a line.
Hey
So my brother posted this on Yahoo
Draw a line from the center of the circle to one of the ends of the chord (water surface) and another to the point at greatest depth. A right-angled triangle is formed. Length of side to the water-surface is 5 cm, the hypot is 7 cm.
<span>What you do now is the following: </span>
<span>Calculate the angle θ in the corner of the right-angled triangle by: cos θ = 5/7 ⇒ θ = cos ˉ¹ (5/7) </span>
<span>So θ is approx 44.4°, so the angle subtended at the center of the circle by the water surface is roughly 88.8° </span>
<span>The area shaded will then be the area of the sector minus the area of the triangle above the water in your diagram. </span>
<span>Shaded area ≃ 88.8/360*area of circle - ½*7*7*sin88.8° </span>
<span>= 88.8/360*π*7² - 24.5*sin 88.8° </span>
<span>≃ 13.5 cm² </span>
<span>(using area of ∆ = ½.a.b.sin C for the triangle) </span>
<span>b) </span>
<span>volume of water = cross-sectional area * length </span>
<span>≃ 13.5 * 30 cm³ </span>
<span>≃ 404 cm³</span>
Hoped it Helped