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1 year ago

Multi step equation -3x-2=-1

Mathematics

1 answer:

Shkiper50 [21]1 year ago

8 0

x = -1/3

Explanation:

-3x-2=-1

collect like terms by adding +2 to both sides:

-3x -2 + 2 = -1 + 2

-3x = 1

Divide both sides by the coefficient of x:

coefficient of x = -3

-3x/-3 = 1/-3

x = -1/3

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I need help on this please I will mark brainliest to first correct answer

viktelen [127]

Answer:

The answer is C. (1, -3)

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3 years ago

A certain number exceeds 18 by as much as 44 exceed the number. What is the number?

asambeis [7]

Answer:

x=31 is the required number.

Step-by-step explanation:

Let the number be represented as "x".

Then, According to the statement given, "x" exceeds by as much as 44 exceed the number, then the equation can be written as:

⇒ $x-18=44-x$

⇒ $x+x=44+18$

⇒ $2x=62$

⇒ $x=31$

Hence, the required number is x =31.

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3 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 $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

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3 years ago

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Rom4ik [11]

The G.C.F of the given algebraic expression is; ⁴⁹/₆a xy

<h3>What is the G.C.F (Greatest Common Factor)?</h3>

The greatest common factor (GCF or GCD or HCF) of a set of whole numbers is the largest positive integer that divides evenly into all numbers with zero remainder. For example, for the set of numbers 18, 30 and 42 the GCF = 6.

We are given the algebraic expression;

(3 x y * 4x²y * ¹/₁ * 5x³y ^ z * 4)7a * 1/1 * 7*6

Expanding this further gives;

⁴⁹/₆a((3xy * 4x²y * 20x³y ^ z )

Now, the GCF of the terms inside the bracket would be x y. Thus, expanding the GCF we have;

⁴⁹/₆a xy((3 * 4x * 20x²y^(z - 1))

Read more about G.C.F at; brainly.com/question/219464

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Serggg [28]

7b is just 7 x b and if b is 6 its just 7 x 6
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3 years ago

Read 2 more answers