All categories

3 years ago

Find the height of the triangle 10 points!

Mathematics

2 answers:

Naddik [55]3 years ago

5 0

Answer:

A. 3.4

Step-by-step explanation:

We have been given a graph of a right triangle and we are asked to find the height of our given right triangle.

Since we know that sine relates the opposite side of a right triangle to hypotenuse, so we will use sine to find height of our given triangle.

$\text{Sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{Sin}(20^{o})=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{Sin}(20^{o})=\frac{x}{10}$

$0.342020143326=\frac{x}{10}$

$0.342020143326\times 10=\frac{x}{10}\times 10$

$0.342020143326\times 10=x$

$x=03.42020143326\approx 3.4$

Therefore, the height of our given triangle is 3.4 units and option A is the correct choice.

Scorpion4ik [409]3 years ago

4 0

By the 6 trigonometric ratios in a right triangle

(sine, cosine, tangent, cotangent, secant, cosecant)

we have the following:

$\displaystyle{ sin20^o= \frac{opposite \ side}{hypothenuse}= \frac{x}{10}$

So, $x=10\cdot sin20^o$

sin20° must be given, or can be found using a calculator:

Using the calculator of our pc: Calculator- view:Scientific- press 20 then sin button, we find : sin20°=0.342

$x=10\cdot sin20^o=10\cdot0.342=3.42$

Answer: A)3.4

You might be interested in

Please Help!!

leva [86]

Answer:

0.0069

Step-by-step explanation:

This is a power series problem.

The taylor power series expansion for sin(x) = x - x³/3! + (x^(5))/5! - (x^(7))/7! + (x^(9))/9! .......

Our question says we should use the first 5 terms to find the value of sin(π). Thus;

sin(π) = π - π³/3! + (π^(5))/5! - (π^(7))/7! + (π^(9))/9!

This gives;

π - (π^(3)/6) + (π^(5))/120 - (π^(7))/5040 + (π^(9))/362880 ≈ 0.0069

5 0

3 years ago

PLLLLLSSSS Heeeellppp ill mark brainliest I promisseeeeeeee plsssss

WARRIOR [948]

Answer:

$2\,(x-1)^2=4$

which is the first option in the list of possible answers.

Step-by-step explanation:

Recall that the minimum of a parabola generated by a quadratic expression is at the vertex of the parabola, and the formula for the vertex of a quadratic of the general form:

$y=ax^2+bx+c$

is at $x_{vertex}=\frac{-b}{2\,a}$

For our case, where $a=2\,\,, b=-4\,\,,\,\,and \,\,c=-2$ we have:

$x_{vertex}=\frac{-b}{2\,a}\\x_{vertex}=\frac{4}{2\,*\,2}\\x_{vertex}=1$

And when x = 1, the value of "y" is:

$y(x)=2x^2-4x-2\\y(1)=2(1)^2-4(1)-2\\y(1)=2-6\\y(1)=-4\\y_{vertex}=-4$

Recall now that we can write the quadratic in what is called: "vertex form" using the coordinates $(x_{vertex},y_{vertex)$ of the vertex as follows:

$y-y_{vertex}=a\,(x-x_{vertex})^2$

Then, for our case:

$y-(-4)=2\,(x-1)^2\\y=2\,(x-1)^2-4$

Then, for the quadratic equal to zero as requested in the problem, we have:

$y=2\,(x-1)^2-4=0\\2\,(x-1)^2-4=0\\2\,(x-1)^2=4$

5 0

3 years ago

Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7

lilavasa [31]

Answer:

<h2>A.Vertical:x=7</h2><h2>Slant:y=x+9</h2>

Step-by-step explanation:

$f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}$

$slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1$

$b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}$