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3 years ago

Which pair of angles are supplementary?

Mathematics

1 answer:

Lina20 [59]3 years ago

3 0

Supplementary angles are angles that add up to 180 degrees, aka one straight line. The supplementary angles here are:

1 and 2, 1 and 3, 2 and 4, 3 and 4, 5 and 6, 6 and 8, 7 and 8, and 6 and 7. Well, those are the more obvious ones. 1 and 7 are supplementary because if you envision them next to each other, you’ll see that they create a straight line. So, with that logic, 3 and 5 are supplementary because when you put them together, they create a straight line

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Factor Completely 14bx^2-7x^3-4b+2x

Mnenie [13.5K]

Answer:

=14bx2−7x3−4b+2x

Step-by-step explanation:

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3 years ago

Read 2 more answers

Guys, please help me with this question.

Elodia [21]

Answer:

Step-by-step explanation:

We are asked to perform the subtraction

x² + xy - 3y² - (5x² - xy + y²) ← distribute the parenthesis by - 1

= x² + xy - 3y² - 5x² + xy - y² ← collect like terms

= (x² - 5x²) + (xy + xy) + (- 3y² - y² )

= - 4x² + 2xy - 4y²

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3 years ago

How do i solve this quadratic equation x²+6x+c

Anuta_ua [19.1K]

recall in a perfect square trinomial, the middle term is the tale-tell guy, we know the middle term is the product of 2 and the two other guys without the exponent, so in this case 6x = 2*√x² * √c.

$\bf 6x=2\cdot \sqrt{x^2}\cdot \sqrt{c}\implies 6x=2\sqrt{x^2c}\implies 6x=2x\sqrt{c}\\\\\\\cfrac{6x}{2x}=\sqrt{c}\implies 3=\sqrt{c}\implies 3^2=c\implies 9=c$

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4 years ago

The number of customers waiting for gift-wrap service at a department store is an rv X with possible values 0, 1, 2, 3, 4 and co

julia-pushkina [17]

Answer:

P[X=3,Y=3] = 0.0416

Step-by-step explanation:

Solution:

- X is the RV denoting the no. of customers in line.

- Y is the sum of Customers C.

- Where no. of Customers C's to be summed is equal to the X value.

- Since both events are independent we have:

P[X=3,Y=3] = P[X=3]*P[Y=3/X=3]

P[X=3].P[Y=3/X=3] = P[X=3]*P[C1+C2+C3=3/X=3]

P[X=3]*P[C1+C2+C3=3/X=3] = P[X=3]*P[C1=1,C2=1,C3=1]

P[X=3]*P[C1=1,C2=1,C3=1] = P[X=3]*(P[C=1]^3)

- Thus, we have:

P[X=3,Y=3] = P[X=3]*(P[C=1]^3) = 0.25*(0.55)^3

P[X=3,Y=3] = 0.0416

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3 years ago

A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest

faltersainse [42]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73$

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396$

add both amounts, and that's her investment for the 11 years.

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3 years ago