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3 years ago

Greatest common factor for 60x4y7, 45x5y5, and 75x3y

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

4 0

Wow.... College is hard...???

Katen [24]3 years ago

3 0

Answer:

the answer is B

Step-by-step explanation:

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Find the missing side lengths. Leave your answers as radicals in simplest form.

steposvetlana [31]

The length of missing sides x = 5 units and y = 5 units.

<h3>What is Trigonometric ratios?</h3><h3>The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec).</h3>

In ΔABC,

AB (Base) = y , BC(Hypotenuse) = $5\sqrt{2}$ , CA (perpendicular) = x

and ∠ABC = $45^0$

Now,

tan $45^0$ = perp. / base

1 = x /y

x= y .................(i)

again,

sin $45^0$ = perp. / Hypo.

$\frac{1}{\sqrt{2} } = \frac{x}{5\sqrt{2} }$

x = 5

put in equation (i), we get

y = 5

Thus, the length of missing side of the given triangle is x = 5 units and y = 5 units.

Learn more about trigonometric ratios from :

brainly.com/question/1201366

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6 0

2 years ago

Calcule o determinate : <br><br> -3 2<br> 5 -3

Darina [25.2K]

The answer to this question is ....................................

4 0

3 years ago

Simplify using the vertical method. (3k + 4)(3k^2 – 5k – 3)

liubo4ka [24]

Answer:

$9k^{3} -3k^{2}-29k-12$

Step-by-step explanation:

Step 1: Expand by distributing sum groups.

$3k(3k^{2} -5k-3)+4(3k^{2} -5k-3)$

Step 2: Expand by distributing terms.

$9k^{3} -15k^{2} -9k+4(3k^{2} -5k-3)$

Step 3: Expand by distributing terms.

$9k^{3}-15k^{2} -9k+12k^{2} -20k-12$

Step 4: Collect like terms.

$9k^{3}+(-15k^{2} +12k^{2} )+(-9k-20k)-12$

Step 5: Simplify.

$9k^{3} -3k^{2}-29k-12$

3 0

3 years ago

Your company gives you a cell phone plan that allows you 600 minutes to talk per month. So far This month, you have used 330 min

Mars2501 [29]

33 per day . divide 330 and 10

8 0

3 years ago

If the probability of observing at least one car on a highway during any 20-minute time interval is 609/625, then what is the pr

andreyandreev [35.5K]

If the probability of observing at least one car on a highway during any 20-minute time interval is 609/625, then the probability of observing at least one car during any 5-minute time interval is 609/2500

Given The probability of observing at least one car on a highway during any 20 minute time interval is 609/625.

We have to find the probability of observing at least one car during any 5 minute time interval.

Probability is the likeliness of happening an event among all the events possible. It is calculated as number/ total number. Its value lies between 0 and 1.

Probability during 20 minutes interval=609/625

Probability during 1 minute interval=609/625*20

=609/12500

Probability during 5 minute interval=(609/12500)*5

=609/2500

Hence the probability of observing at least one car during any 5 minute time interval is 609/2500.

Learn more about probability at brainly.com/question/24756209

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8 0

2 years ago