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3 years ago

6

A rectangular cardboard is 65 cm long and 45 cm wide. What is the maximum number of rectangles, each 12 cm long and 8 cm wide th

at can be cut from it?

1 answer:

Galina-37 [17]3 years ago

8 0

The answer is thirty

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A. Sales are (increasing/decreasing)…. _____….(purchases, purchases/month, months/purchases, $/purchases, purchases/$, $, months

lord [1]

Answer:

Sales are (increasing/decreasing)…. _____….(purchases, purchases/month, months/purchases, $/purchases, purchases/$, $, months)

D. (January, February, March, April, May, June, July, August, September, October, November, December)

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2 years ago

Use complete sentences to describe why postulates are helpful?

Savatey [412]

<span>Postulates are helpful because it helps you plan and solve geometry problems.
That's my guess. It doesn't tell me in the text either.</span>

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2 years ago

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Two boats sail away from a buoy making a 90 degree angle. The first boat is 165 meters away and is traveling at 25 m/s. The seco

NikAS [45]

Answer:

Step-by-step explanation:

lets say in the shown triangle x=165

so speed $\frac{dx}{dt} =25$

and $\frac{dy}{dt} =35$

Also since distance = speed* time

time = pi minutes = pi/60 seconds

y = $\frac{35\pi \pi }{60}$

y = $\frac{7\pi }{12}$

Now using the pythagorean :

$x^2+y^2=z^2\\so \\165^2+(\frac{7\pi }{12} )^2 =z^2$

<u>derivate the equation so we get :</u>

$x\frac{dx}{dt} +y\frac{dy}{dt} =z\frac{dz}{dt} \\$ $165*25+35*\frac{7\pi }{12}=\sqrt{165^2+(\frac{7\pi }{12})^2} *\frac{dz}{dt}$

$\frac{dz}{dt} =25.387$ m/s

So Rate of change = <u><em>25.387 m/s</em></u>

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3 years ago

What is true regarding two adjacent arcs created by two intersecting diameters?

ozzi

Answer – The sum of their measures is 180 degreed

When two diameters of the same circle intersect, four arcs are created. The
following are true of these arcs:
- Opposite arcs are equal in measure
- The sum of the measure of adjacent arcs is 180 degrees
- The sum of the measure of all the arcs is 360 degrees

6 0

3 years ago

Read 2 more answers

(1/(x-1))-(3/(x^2+2x-3))<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bx%20-%201%7D%20%20-%20%20%5Cfrac%7B3%7D%7Bx%20

ycow [4]

$\dfrac{1}{x-1}-\dfrac{3}{x^2+2x-3} \\\dfrac{1}{x-1}-\dfrac{3}{(x+3)(x-1)} \\\dfrac{1\cdot(x+3)-3}{(x+3)(x-1)} \\\dfrac{x+3-3}{(x+3)(x-1)} \\\dfrac{x}{(x+3)(x-1)}$

Hope this helps.

r3t40

4 0

3 years ago