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UNO [17]
3 years ago
12

What is the greatest common factor of 15 and 65

Mathematics
2 answers:
svet-max [94.6K]3 years ago
8 0

Answer: 5


Step-by-step explanation:

The most you can divide by 15 and 65 to get a whole number would be 5

miskamm [114]3 years ago
6 0
The factors of 15 are 1,3,5,15
The factors of 65 are 1,5,13,65

the greatest common factor is 5
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(-2/3, sqrt5/3) is a point on a unit circle. Find the cosine, cosecant, and sine of the angle.
Nana76 [90]

Look at the picture.

\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{\frac{y}{r}}=\dfrac{r}{y}

We have the right triangle x, y and r. From the Pythagorean theorem we have:

r^2=x^2+y^2\to r=\sqrt{x^2+y^2}

We have the point

\left(-\dfrac{2}{3};\ \dfrac{\sqrt5}{3}\right)

Substitute:

r=\sqrt{\left(-\dfrac{2}{3}\right)^2+\left(\dfrac{\sqrt5}{3}\right)^2}\\\\r=\sqrt{\dfrac{4}{9}+\dfrac{5}{9}}\\\\r=\sqrt{\dfrac{9}{9}}\\\\r=1

\csc\theta=\dfrac{1}{\frac{\sqrt5}{3}}=\dfrac{3}{\sqrt5}=\dfrac{3\cdot\sqrt5}{\sqrt5\cdot\sqrt5}=\boxed{\dfrac{3\sqrt5}{5}}

8 0
3 years ago
How do you determine the slope interscept equation of two given pointa
Airida [17]
Choose your points, go up from the y intercept to a chosen x point. Then go over the the x point. Up and over is the formula I use.
4 0
3 years ago
How do you solve for y in the equation: <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7By%7D%20%3Dy-6" id="TexFormula1" title
kozerog [31]

No it isn't.

Explanation:

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x = y^2 - 6y

A function gives just one y for every x

In this case there will always be 2 y's for every x

Example:

y can be

y = 6

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(0,-6) & (0,6)

5 0
3 years ago
Please help will mark brainliest
Levart [38]

Answer:

14

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Which expression equals 102+84
MakcuM [25]

Answer: 6(17+14) or 186

Step-by-step explanation: I provide with you two answers since not sure what u  mean .And the first answer is distributive it and the second answer is the sum.

Also 102 + 84 is an expression already so I don't know what your asking and if this help  u are welcome !

4 0
2 years ago
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