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3 years ago

11

If quadrilateral WXYZ is inscribed in a circle with center O, the measure of angle W = 45 and the measure of angle X = 110, then

the measure of angle Z = _____.

1 answer:

velikii [3]3 years ago

7 0

Check the picture below.

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HELP ME PLEASEEEEEEEEEE

Anastasy [175]

Answer:

I can't see the picture sorry

Step-by-step explanation:

4 0

3 years ago

I don’t know how to solve this

Dmitrij [34]

Answer:

y = $\frac{1}{4}$ x - 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) =( 8, - 3) and (x₂, y₂ ) = (0, - 5)

m = $\frac{-5+3}{0-8}$ = $\frac{-2}{-8}$ = $\frac{1}{4}$

Note the line crosses the y- axis at (0, - 5 ) ⇒ c = - 5

y = $\frac{1}{4}$ x - 5 ← equation of line

4 0

3 years ago

Read 2 more answers

Graphs of what functions are shown below?

shtirl [24]

Answer:

-3/5

Step-by-step explanation:

6 0

3 years ago

Lily took out a 5-year loan from the bank for $31,000 to purchase a new car. At the end of the loan, she had paid a total of $37

otez555 [7]

Answer:

=24.5%

Step-by-step explanation:

Simple interest = (principal×rate×time)÷100. *brackets first*
transpose the formula to make rate the subject: rate= (100×simple interest) ÷ (principal×time)
plug in values: rate = (100×37975) ÷ (31000×5)
the result is 24.5%

8 0

3 years ago

How to solve logarithmic equations as such

Serga [27]

$\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5$

$\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}$

5 0

3 years ago