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stiv31 [10]
2 years ago
15

A sequence of the form a1, a1+d, a1+2d, a1+3d, . . . is called a/an _____.

Mathematics
1 answer:
saw5 [17]2 years ago
7 0

Answer:

arithmetic sequence

Step-by-step explanation:

please mark as brainlist

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Graph.<br><br> y+4=2/5(x−3)<br> plz help
Dominik [7]

Answer:

Step-by-step explanation:

Use DESMOS to make the graph.  You can either rearrange the equation to put it in the form y = mx + b, or leave it as written.  The attached graph, has bot, and they overlap.

y+4=2/5(x−3)

or

y = (2/5)x-5.2

5 0
3 years ago
Hi everyone , Would you help me please :
Veseljchak [2.6K]
 
\displaystyle  \\ &#10; \frac{1}{ \sqrt{2} } = \frac{1 \times \sqrt{2}}{ \sqrt{2} \times \sqrt{2}} = \boxed{\frac{\sqrt{2}}{2}}



8 0
3 years ago
A lidless box is to be made using 2m^2 of cardboard find the dimensions of the box that requires the least amount of cardboard
Jlenok [28]
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3 . Find the dimensions of the box that requires the least amount of cardboard. Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From these, we obtain x 2y = 8 = xy2 . This forces x = y = 2, which forces z = 1. Calculating second derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus, moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
5 0
3 years ago
Please help I will give brainlest or whatever!!
Mashutka [201]
The answer is D 6z+42
8 0
2 years ago
Someone plz help me<br><br><br> .
Alex787 [66]

Answer:

12, 4, 24, 30, and 14 are the answers

5 0
2 years ago
Read 2 more answers
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