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Olegator [25]
3 years ago
10

Find the missing angle on a quadrilateral. Angle C equals 80 degrees, Angle D equals 90 degrees, and angle E equals 135 degrees.

What is the measure of Angle F?
Mathematics
1 answer:
padilas [110]3 years ago
4 0
All quadrilateral angles have to add up to a total of 360°, so we just need to add the angles we already have and subtract that from 360.
135+90+80=305
360-305=55 so your answer is 55!
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Nth sequence term question
SSSSS [86.1K]
One example is when n is 11. 11^2 + 4 = 125. 11 is odd but the 11th term of the sequence, 125, is not prime because 125 is divisible by 5 and 25.
6 0
3 years ago
The graph shows the solution to which system of equations?
vichka [17]
Start by seeing where the lines go. They cross (intersect) at (1, -1). We can use this to check later.
Now for slope: green is -4/2 = -2
pink is 4/2 = 2
Now we can create the equations
Let's make green = g(x) and pink = p(x)
if we use y = mx + b, then the green has a y-intercept (b) of +1
So g(x) = -2x + 1
pink has a y-intercept of -3, so p(x) = 2x - 3
Now let's plug n play: put our solutions x into each equation and confirm that it makes the y = -1
g(x) = -2x + 1 = -2(1) + 1 = -2+1 = -1
p(x) = 2x - 3 = 2(1) - 3 = 2-3 = -1

✔ YES THEY ARE CONFIRMED
6 0
3 years ago
Read 2 more answers
I need help this is hard please help me
Mademuasel [1]

Answer:

P(t)=25000+1.12t

Step-by-step explanation:

Now can you help me?

A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following function:

f(t) = −16t2 + 16t + 32

Which of the following is a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground?

3 0
3 years ago
Read 2 more answers
Please help me solve this problem ASAP
Andru [333]
Quadrant 4
Hope this helps!
4 0
2 years ago
Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.
Hitman42 [59]
\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\

\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}

\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
\\\\\\
\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}

7 0
3 years ago
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