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3 years ago

A triangle has side lengths of 15, 20, and 25. What type of triangle is it?

Mathematics

2 answers:

Mrac [35]3 years ago

6 0

Answer:

The answer is a. right.

Step-by-step explanation:

GIVEN:

hypo=25

opposite and adjacent=15 and 20.

FORMULA:

hypotenuse^2 =opposite side^2 + adjacent side^2

25^2=15^2+20^2

625=225+400

400+225=625

(Got this from yahoo answers, everybody said right.)

Vinvika [58]3 years ago

3 0

Acute triangle is the answer

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In 1985, there were 285 cell phone subscribers in Arlington. The number of subscribers increased by 75% per year since 1985. Fin

ANTONII [103]

Answer:

The number of subscribers in 2008 is 110,845,988

Step-by-step explanation:

The number of cell phone subscribers in Arlington, in t years after 1985, can be modeled by the following equation.

$N(t) = N(0)(1+r)^{t}$

And which N(0) is the number of subscribers in 1985 and r is the rate it increases per year, as a decimal.

In 1985, there were 285 cell phone subscribers in Arlington.

This means that $N(0) = 285$

The number of subscribers increased by 75% per year since 1985.

This means that $r = 0.75$

$N(t) = 285(1.75)^{t}$

Find the number of subscribers in 2008.

2008 is 2008-1985 = 23 years after 1985. So we have to find N(23).

$N(23) = 285(1.75)^{23} = 110,845,988$

The number of subscribers in 2008 is 110,845,988

8 0

3 years ago

Find the optimal solution for the following problem. (Round your answers to 3 decimal places.)

Sphinxa [80]

Answer:

x=2.125

y=0

C=19.125

Step-by-step explanation:

To solve this problem we can use a graphical method, we start first noticing the restrictions $x\geq 0$ and $y\geq 0$ , which restricts the solution to be in the positive quadrant. Then we plot the first restriction $8x+10y\leq 17$ shown in purple, then we can plot the second one $11x+12y\leq 25$ shown in the second plot in green.

The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.

So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.