Can someone help me my class is at 9

Mathematics

1 answer:

ololo11 [35]2 years ago

6 0

Answer: 1/4

Step-by-step explanation: 20 is a fourth of 80

Please mark brainleist if I got it right. I really need it to level up

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What is 2x-y=6 the slope

Flura [38]

First of all, lets get y on its own side of the equation.

$-y=-2x+6\\y=2x-6\\$

This is slope intercept form, y=mx+b, where m is slope.

we can see that m=2, so the slope is 2 or $\frac{2}{1}$

5 0

3 years ago

Read 2 more answers

A right triangle has one vertex on the graph of y = 9 - x^2 , x > 0, at ( x , y ), another at the origin, and the third on th

jenyasd209 [6]

Answer:

a. A(x) = (1/2)x(9 -x^2)

b. x > 0 . . . or . . . 0 < x < 3 (see below)

c. A(2) = 5

d. x = √3; A(√3) = 3√3

Step-by-step explanation:

a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...

A(x) = (1/2)(x)(y)

A(x) = (1/2)(x)(9-x^2)

b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).

On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).

c. A(2) = (1/2)(2)(9 -2^2) = 5

The area is 5 when x=2.

d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).