I need this done fast, it's due today!

Mathematics

2 answers:

Bogdan [553]3 years ago

8 0

Answer:

(0, 50) (25, 57.5) (50, 122.5) (75, 195) (100, 275)

Step-by-step explanation:

We know that he earns 30% percent of the money made on sales, plus $50 a week. We know that he starts off with 50, because that's how much he made by working that week. To figure out how much money he made off of each one, we have to calculate 30% of the sale, and add that to the total amount of money he has made up to that point. I hope this helps :)

alexira [117]3 years ago

6 0

Answer:

y=50x(.3)+30

Step-by-step explanation:

You get an average of $50 per week and then work 30 hours per week but the outcome is multiplied by .30 from the sales commision so the answer would be y=50x(.3)+30

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Someone help, lol I've been tryna do the last too for so long, I think I am just tired...

kozerog [31]

Answer:

see explanation

Step-by-step explanation:

(a)

4 × 4 = 4²

(b)

5 × 5 × 5 × 5 = $5^{4}$

(c)

3 × 7 × 7 × 7 × 7 = 3 × $7^{4}$

(d)

2 × 9 × 9 × 9 × 2 × 9 ← rearrange

= 2 × 2 × 9 × 9 × 9 × 9

= 2² × $9^{4}$

6 0

3 years ago

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).