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Dvinal [7]
4 years ago
10

What is the sixth term in the addition pattern that begins with 15,29,43,57?

Mathematics
1 answer:
WITCHER [35]4 years ago
5 0

Answer:

C is your answer

Step-by-step explanation:

Add every number by 14

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A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation
Alexxx [7]
Ok

y=-0.04x^2+8.3x+4.3
when the rocket reaches the ground (when height=0, ie when y=0), then the rocket will land, find the x coordinate

set y=0
0=-0.04x^2+8.3x+4.3
use quadratic formula
if you have ax^2+bx+c=0, then
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}
a=-0.04
b=8.3
c=4.3
x=\frac{-8.3+/- \sqrt{8.3^{2}-4(-0.04)(8.3)} }{2(-0.04)}
x=208.017 or -0.516785
xrepresents horizontal distance
you cannot have a negative horizontal distance unless you fired and theh wind blew it backwards
therefor x=280.017 is the answer
208.02 m


7 0
3 years ago
I don’t understand geometry
Talja [164]

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5 0
4 years ago
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Wich expression can be used to find the slope of a line containing the points (-3,2) and (7,-1)
docker41 [41]
A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\\text{The slope:}\ m=\dfrac{y_2-y_1}{x_2-x_1}


\text{We have:}\\(-3;\ 2)\to\ x_1=-3;\ y_1=2\\(7;\ -1)\to x_2=7;\ y_2=-1

\text{substitute}\\\\m=\dfrac{-1-2}{7-(-3)}=\dfrac{-3}{10}=-\dfrac{3}{10}
6 0
4 years ago
Element X decays radioactively with a half-life of eight minutes. If there are 800 g of element X, how long, to the nearest 10th
Harman [31]

Answer: 41.5 min

Step-by-step explanation:

This problem can be solved with the Radioactive Half Life Formula:  

A=A_{o}.2^{\frac{-t}{h}} (1)

Where:  

A=22 g is the final amount of the radioactive element

A_{o}=800 g is the initial amount of the radioactive element  

t is the time elapsed  

h=8 min is the half life of the radioactive element  

So, we need to substitute the given values and find t from (1):  

22 g=(800 g) 2^{\frac{-t}{8 min}} (2)  

\frac{22 g}{800 g}=2^{\frac{-t}{8 min}} (3)  

\frac{11}{400}=2^{\frac{-t}{8 min}} (4)  

Applying natural logarithm in both sides:  

ln(\frac{11}{400})=ln(2^{\frac{-t}{8 min}}) (5)  

-3.593=-\frac{t}{8 min}ln(2) (6)  

Clearing t:

t=41.46 min \approx 41.5 min This is the time elapsed

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3 years ago
Which of the following would be an appropriate horizontal and vertical scale of a viewing window on a graphing calculator for th
Tpy6a [65]
The eee part is the right part because I’m not trying to get points I’m so sorry
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