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3 years ago

(4) need help and the answer

Mathematics

1 answer:

Oksi-84 [34.3K]3 years ago

3 0

OK well it would be A But then it would be C but A & C Are Wrong so it Would be D

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Mel received notice that his monthly health care rate will increase 20% next year. If Mel currently pays $150 per month for heal

sveta [45]

It will increase by $30

We need to convert the percentage to a decimal by dividing by 100. Then we multiply by the $150.

150 x 0.20 = 30

7 0

3 years ago

What Properties or characteristics or similar triangles could be used to prove the Pythagorean theorem?

Zolol [24]

Well, in order to prove the Pythagorean theorem, every triangle you are using has to have one right angle (90-degree angle), and the side opposite to it will be called the hypothenuse. The remaining two triangles will be acute angles (<90 degrees), and the sides opposite to them are called sides/catheti.

4 0

3 years ago

Find the max value of c=5x+4y

Zigmanuir [339]

You should be able to find the maximum value of c by simply substituting the maximum values for x and y.

So we'll let x = -2 and y = 6.

The equation is is now as follows:

c = 5(-2) + 4(6)

c = -10 + 24

c = 14

Therefore the maximum value for c is 14.

6 0

3 years ago

After plotting the data where x represents the length of the side of a polygon, and f(x) represents the area of the polygon, Jac

garri49 [273]

Answer:

C. 19

Step-by-step explanation:

$f(x) = x {}^{2} + 3x + 1 \\ f(3) = 3 {}^{2} + 3(3) + 1 \\ = 19$

7 0

3 years ago

Find the area of the region in the first quadrant bounded on the left by the y-axis, below by the line y equals one third x co

bonufazy [111]

Answer:

The bounded area is: $\frac{73}{6}\approx 12.17$

Step-by-step explanation:

Let's start by plotting the functions that enclose the area, so we can find how to practically use integration. Please see attached image where the area in question has been highlighted in light green. The important points that define where the integrations should be performed are also identified with dots in darker green color. These two important points are: (2, 6) and (3, 1)

So we need to perform two separate integrals and add the appropriate areas at the end. The first integral is that of the difference of function y=x+4 minus function y=(1/3)x , and this integral should go from x = 0 to x = 2 (see the bottom left image with the area in red:

$\int\limits^2_0 {x+4-\frac{x}{3} } \, dx =\int\limits^2_0 {\frac{2x}{3} +4} \, dx=\frac{4}{3} +8= \frac{28}{3}$

The next integral is that of the difference between $y=-x^2+10$ and the bottom line defined by: y = (1/3) x. This integration is in between x = 2 and x = 3 (see bottom right image with the area in red:

$\int\limits^3_2 {-x^2+10-\frac{x}{3} } \, dx =-9+30-\frac{3}{2} -(-\frac{8}{3} +20-\frac{2}{3} )=\frac{39}{2} -\frac{50}{3} =\frac{17}{6}$

Now we need to add the two areas found in order to get the total area:

$\frac{28}{3} +\frac{17}{6} =\frac{73}{6}\approx 12.17$

8 0

4 years ago