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2 years ago

ZX and zy are supplementary angles. zy measures 33°.what is the measure of ZX

Mathematics

1 answer:

Sati [7]2 years ago

8 0

Answer:

147°

Step-by-step explanation:

supplementary angles equal to 180° in total so...

180 - 33 = 147

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WILL MARK BRAINLIEST!! Which of the following ordered pairs are y-intercepts? Check all that apply.

Y_Kistochka [10]

Y intercept is where the line meets the Y axis
So the points shd be in (0,y) form.

So (0,0) (0,-7) (0,-0.25) are the y-intercepts in the following.

8 0

2 years ago

Read 2 more answers

Solve this system of equations:<br><br> 5x + y = 7<br> 20x + 2 = y

xz_007 [3.2K]

Answer:

x = 1/5, y = 6

Step-by-step explanation:

5x + y = 7

y = 7 - 5x

Set equations equal to each other to eliminate y

7 - 5x = 20x + 2

7 = 25x + 2

5 = 25x

x = 1/5

y = 7 - 5(1/5)

y = 7 - 1

y = 6

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2 years ago

5x = -25 what is the answer of x

Sliva [168]

Answer:

x= -5

Step-by-step explanation:

just use a calculator

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2 years ago

At sunrise, the temperature was -12 °F. The temperature increased 19 °F each hour for 3 hours. What was the temperature 3 hours

yawa3891 [41]

-12 degree F + 19 + 19 + 19 = 49 degree F
I think the answer is 49 degree F...

4 0

2 years ago

Businesses deposit large sums of money into bank accountsImagine an account with $10 million dollars in it.

adoni [48]

again, let's assume daily compounding means 365 days per year.

$~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ t=years\dotfill &1 \end{cases} \\\\\\ I = (10000000)(0.0212)(1)\implies \boxed{I=212000}$

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10000000\\ r=rate\to 2.12\%\to \frac{2.12}{100}\dotfill &0.0212\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\dotfill &365\\ t=years\dotfill &1 \end{cases}$

$A=10000000\left(1+\frac{0.0212}{365}\right)^{365\cdot 1}\implies A\approx 10214256.88 \\\\\\ \underset{\textit{earned interest amount}}{10214256.88~~ - ~~10000000 ~~ \approx ~~ \boxed{214256.88}}$

what's their difference? well

$\stackrel{\textit{compounded daily}}{\approx 214256.88}~~ - ~~\stackrel{\textit{simple interest}}{212000}\implies \boxed{2256.88}$

7 0

1 year ago