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ehidna [41]
3 years ago
12

Write the sum using summation notation, assuming the suggested pattern continues. 2 - 6 + 18 - 54 + ...

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
8 0
well for that part of the equation : the ans. is -40, but what bout' the other part ?

vlada-n [284]3 years ago
5 0
<span>E with infinity over it and n=0 under it with 2(-3)^n to the side</span>
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On the grid draw the graph of y= 1- 3x for values of x from -2 to 3
tino4ka555 [31]

Answer:

see solution

Step-by-step explanation:

equation: y = 1 - 3x

when x is -2:
y = 1 -3 * -2  = y = 1+6 = y = 7
coordinates = (x,y) = (-2,7)

when x is -1:
y = 1-3*-1 = y = 1+3 = y = 4
coordinates = (x,y) = (-1,4)

when x is 0:
y = 1 - 3 * 0 = y = 1-0 = y =1
coordinates = (x,y) = (0,1)

when x is 1
y = 1-3*1 = y = 1-3 = y = -2
coordinates = (x,y) = (1,-2)

When x is 2
y = 1-3*2 = y = 1-6 = y = -5
coordinates = (x,y) = (2,-5)

when x is 3
y = 1-3*3 = y = 1-9 = y = -8
coordinates = (x,y) = (3,-8)

Plot these points on the graph and join up the dots to see the line.

bye

5 0
1 year ago
Cuál es el resultado de 1/3 + 5/6?​
Sladkaya [172]
El resultado es 7/6
5 0
2 years ago
Read 2 more answers
Evaluate triple integral ​
kaheart [24]

Answer:

\\ \frac{1}{8} e^{4a}-\frac{3}{4}e^{2a}+e^{a} -\frac{3}{8} \\\\or\\\\ \frac{e^{4a}-6e^{2a}+8e^{a}-3}{8}

Step-by-step explanation:

\\ \int\limits^{a}_{0} \int\limits^{x}_{0} \int\limits^{x+y}_{0} {e^{x+y+z}} \, dzdydx \\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [\int\limits^{x+y}_{0} {e^{x+y}e^z} \, dz]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}\int\limits^{x+y}_{0} {e^z} \, dz]dydx\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^z\Big|_0^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} [e^{x+y}e^{x+y}-e^{x+y}]dydx \\\\\\=\int\limits^{a}_{0} \int\limits^{x}_{0} e^{2x+2y}-e^{x+y}dydx \\\\\\

\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}-e^{x+y}dy]dx \\\\\\=\int\limits^{a}_{0} [\int\limits^{x}_{0} e^{2x}e^{2y}dy- \int\limits^{x}_{0}e^{x}e^{y}dy]dx \\\\\\u=2y\\du=2dy\\dy=\frac{1}{2}du\\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\int e^{u}du- e^x\int\limits^{x}_{0}e^{y}dy]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x}}{2}\cdot e^{2y}\Big|_0^x- e^xe^{y}\Big|_0^x]dx \\\\\\=\int\limits^{a}_{0} [\frac{e^{2x+2y}}{2} - e^{x+y}\Big|_0^x]dx \\\\

\\=\int\limits^{a}_{0} [\frac{e^{4x}}{2} - e^{2x}-\frac{e^{2x}}{2} + e^{x}]dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2} -\frac{3e^{2x}}{2} + e^{x}dx \\\\\\=\int\limits^{a}_{0} \frac{e^{4x}}{2}dx -\int\limits^{a}_{0}\frac{3e^{2x}}{2}dx + \int\limits^{a}_{0}e^{x}dx \\\\\\u_1=4x\\du_1=4dx\\dx=\frac{1}{4}du_1\\\\\u_2=2x\\du_2=2dx\\dx=\frac{1}{2}du_2\\\\\\=\frac{1}{8}\int e^{u_1}du_1 -\frac{3}{4}\int e^{u_2}du_2 + \int\limits^{a}_{0}e^{x}dx \\\\\\

\\=\frac{1}{8}e^{u_1}\Big| -\frac{3}{4}e^{u_2}\Big| + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x}\Big|_{0}^a -\frac{3}{4}e^{2x}\Big|_{0}^a + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4x} -\frac{3}{4}e^{2x} + e^{x}\Big|_0^a \\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{1}{8} +\frac{3}{4} -1\\\\\\=\frac{1}{8}e^{4a} -\frac{3}{4}e^{2a} + e^{a}-\frac{3}{8}\\\\\\

Sorry if that took a while to finish. I am in AP Calculus BC and that was my first time evaluating a triple integral. You will see some integrals and evaluation signs with blank upper and lower boundaries. I just had my equation in terms of u and didn't want to get any variables confused. Hope this helps you. If you have any questions let me know. Have a nice night.

6 0
2 years ago
Pls help me
aivan3 [116]
The and answer is going to be 3 because when you multiply 5 and 3 you get 15
3 0
2 years ago
A rectangle has a perimeter of 26 cm and one of its sides has a length of 5 cm sketch a rectangle and label all of its sides len
vovikov84 [41]
Im thinking both shorter sides are 5cm, and the 2 longer sides are 8cm each... add them all together and you get 26cm
5 0
3 years ago
Read 2 more answers
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