Please help me label the diagram beaded on the word bank given. (Geometry)

HELPPPPPPPPPPPPP MEEEEEEEEEEEEEEEEEEE

Kryger [21]

Answer:

132°

Step-by-step explanation:

Angle JKH and IHK are alternate interior angles, and thus congruent.

Since JKH is 132 degrees, so is IHK

7 0

3 years ago

Can u guys answer these last 4 questions for me w/ work shown and also label the ones u guys did so I won’t get them mixed up

bagirrra123 [75]

Answer:

it's simple substitution

Step-by-step explanation:

1. (7)(2)/2= 14/2=7

2. (5)-(8)/4=

3. (6)-(1)+6=11

4. (19)+(1)^3=21

4 0

3 years ago

A particle moves in a straight line so that its velocity at time

riadik2000 [5.3K]

Answer:

s(2) = 7.75

Step-by-step explanation:

given the velocity v(t) = t^3

we can find the position s(t) by simply integrating v(t) and using the boundary conditions s(1)=2

$s(t) = \int {v(t)} \, dt\\ s(t) = \int {t^3} \, dt\\s(t) = \dfrac{t^4}{4}+c$

we know throught s(1) = 2, that at t=1, s =2. we can use this to find the value of the constant c.

$s(1) = \dfrac{1^4}{4}+c\\4 = \dfrac{1^4}{4}+c\\c = 4-\dfrac{1}{4}\\c = \dfrac{15}{4} = 3.75$

Now we can use this value of t to formulate the position function s(t):

$s(t) = \dfrac{t^4}{4}+\dfrac{15}{4}\\$

this is the position at time t.

to find the position at t=2

$s(2) = \dfrac{2^4}{4}+\dfrac{15}{4}\\$

$s(2) = \dfrac{31}{4} = 7.75$

the position of the particle at time, t =2 is s(2) = 7.75

3 0

3 years ago

Write the Recursive rule for the geometric sequence<br>An=1072-1<br>8, 4, 2, 1, 1/2,...

PIT_PIT [208]

$\bf 8~~,~~\stackrel{8\cdot \frac{1}{2}}{4}~~,~~\stackrel{4\cdot \frac{1}{2}}{2}~~,~~\stackrel{2\cdot \frac{1}{2}}{1}~~,~~\stackrel{1\cdot \frac{1}{2}}{\cfrac{1}{2}} \\\\\\ a_n=\cfrac{1}{2}\cdot a_{n-1}\qquad \begin{cases} a_1=\textit{previous term}\\ a_n=\textit{current term}\\ a_1=\textit{first term}\\ \qquad 8 \end{cases}$

4 0

3 years ago

If you were to deposit $12,000 and the bank paid you 6% interest compounded quarterly, what would the maturity value of your dep

xenn [34]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{compounded amount}\\
P=\textit{original amount deposited}\to &\$12000\\
r=rate\to6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, meaning}
\end{array}\to &4\\

t=years\to &9
\end{cases}
\\\\\\
A=12000\left(1+\frac{0.06}{4}\right)^{4\cdot 9}$

4 0

3 years ago