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2 years ago

What is the value of one 1/2 Divided by 1/3

Mathematics

2 answers:

Natalka [10]2 years ago

8 0

the answer is 1/6
1/2÷1/3=1/6

olga nikolaevna [1]2 years ago

7 0

1/2(1/3)= 0.166667

Hope this helps!

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On a coordinate plane, triangle A B C is shown. Point A is at (negative 1, 1), point B is at (3, 2), and points C is at (negativ

oksian1 [2.3K]

The corresponding height of the triangle is 1.6 units

The formula for calculating the area of a triangle is expressed as:

$A=\frac{1}{2} bh$

b is the base of the triangle

h is the height of the triangle

Given the coordinates of the base BC of the triangle given as B(3, 2), and C(-1,-1). Using the distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\BC= \sqrt{(-1-2)^2+(-1-3)^2}\\BC=\sqrt{3^2+4^2}\\BC=\sqrt{25}\\BC=5units$

The area of the triangle passing through the coordinate points A(-1, 1), B(3,2), and C(-1, -1) is expressed as:

$A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\$

Substituting the coordinate points:

$A=\frac{1}{2}[(-1)(2-(-1))+3(-1-1)+-1(1-2)]\\A=\frac{1}{2}[(-1)(3)+3(-2)+-1(-1)]\\A=\frac{1}{2}[-3-6+1]\\A=\frac{1}{2} (-8)\\|A| =4 units^2$

Recall that:

$A = 0.5bh\\h=\frac{A}{0.5b}\\h=\frac{4}{0.5(5)}\\h=\frac{4}{2.5}\\h= 1.6 units$

Hence the corresponding height of the triangle is 1.6 units

Learn more on area of triangles here: brainly.com/question/17335144

8 0

2 years ago

Can anyone pls help me I really need help pls help me I will mark them as brainiest pls!!! Can u do it in a passer and then send

Andreas93 [3]

Answer:

Hi there!

I provided the charts down below. Hope this helps! ;)

8 0

2 years ago

Please help me! What is the simplest form for 21 /63 ?

creativ13 [48]

Answer:

1/3

Step-by-step explanation:

5 0

2 years ago

How to know if a function is periodic without graphing it ?

zhenek [66]

A function $f(t)$ is periodic if there is some constant $k$ such that $f(t+k)=f(k)$ for all $t$ in the domain of $f(t)$ . Then $k$ is the "period" of $f(t)$ .

Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

$\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5$

Expanding on the left, you have

$\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2$

and

$1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5$

It follows that the following must be satisfied:

$\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}$

The first two equations are satisfied whenever $k\in\{0,\pm4,\pm8,\ldots\}$ , or more generally, when $k=4n$ and $n\in\mathbb Z$ (i.e. any multiple of 4).

The second two are satisfied whenever $k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}$ , and more generally when $k=\dfrac{10n}7$ with $n\in\mathbb Z$ (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when $k$ is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

$\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2$

$\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5$

More generally, it can be shown that

$f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))$

is periodic with period $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

4 0

2 years ago

Help pls will give brainliest to correct answer

Marysya12 [62]

D is correct answers

7 0

2 years ago