All categories

3 years ago

Solve for x. PLZ HELP ASAP!!!

Mathematics

1 answer:

ycow [4]3 years ago

3 0

X is a vertical angle to the angle marked as 100 degrees.

Vertical angles are the same so x = 100 degrees

Answer: 100 degrees

You might be interested in

A new shopping mall records 120 total shoppers on their first day of business. Each day after that, the number of shoppers is 10

Marta_Voda [28]

Answer:

1138

Step-by-step explanation:

From the information given:

We can represent it perfectly in an exponential form:

$m = p(q)^x$

where;

p = initial value = 120

q = base of the exponential form

q = 1 + r

here; r = rate in decimal = 10% = 0.1

Then q can now be = 1 + 0.1 = 1.1

Replacing it into the exponential form, we get:

$m = 120(1.1)^x$

where;

x = number of days and m = number of shoppers

Thus:

For the first day:

$m = 120(1.1)^x$

m = 120

For the second day:

$m = 120(1.1)^1$

m = 132

For the third day:

$m = 120(1.1)^2$

m = 145.2

For the fourth day

$m = 120(1.1)^3$

m = 159.72

For the fifth-day

$m = 120(1.1)^4$

m = 175.692

For the sixth-day

$m = 120(1.1)^5$

m = 193.2612

For the seventh-day

$m = 120(1.1)^6$

m = 212.58732

Thus; the total numbers of shoppers for the first 7 days is:

$= 120+ 132 + 145.2 + 159.72 + 175.692+193.2612+212.58732$

= 1138.46052

≅ 1138

6 0

3 years ago

Solve using algebraic equation: 5sin2x=3cosx (No exponents)

DaniilM [7]

$5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x$

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .