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3 years ago

An investment grows by 3% per year for 20 years. By what percent does it increase over the 20 year period?

Mathematics

1 answer:

Y_Kistochka [10]3 years ago

7 0

3%?? The question doesn't make much sense.

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Triangle ABC has vertices of A(–8 ,8), B(6, 2), and C(–2, 1). Find the length of the median from C in triangle ABC

babunello [35]

The length of the median from vertex C is equal to √17. As a median of a triangle is a line segment joining a single vertex to the midpoint of the opposite side of the triangle. In this case, the median will be from vertex C to the mid-point of the triangles side AB.<span> Thus, we can work out the length of the median from vertex C by using the Midpoint formula; M(AB) = (X</span>∨1 + X∨2) /2 ; (Y∨1 + Y∨2) /2 . Giving us the points of the midpoint of side AB, which can be plotted on the cartesian plane. to find the length of the median from vertex C, we can use the distance formula and the coordinates of the midpoint and vertex C , d = √(X∨2 - X∨1) ∧2 + (Y∨2 - Y∨1)∧2.

4 0

3 years ago

Okay will award brainliest and put alot of points so pls help I'm desparate!!!

Bad White [126]

Answer:

23 is 31/100

Step-by-step explanation:

not sure about the rest sorry :(

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3 years ago

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What is the period of f(x) = sin(x)? StartFraction pi Over 2 EndFraction Pi StartFraction 3 pi Over 2 EndFraction 2 pi

Nostrana [21]

<h3>Answer: 2pi</h3>

2pi radians are equal to 360 degrees. After this point, the function f(x) = sin(x) repeats its values again. Each cycle is 2pi radians long.

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2 years ago

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

2 years ago

What is the exact circumference of a circle with a radius of 45 cm?<br><br> HELP ASAP!!!

Sergeeva-Olga [200]

The answer is 282.7433

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3 years ago

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