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12345 [234]
2 years ago
7

heeeeeeeeeeeeeeeeeeeeeeeeeeeeellllllllllllllllllllllllllllllllllllllllllppppppppppppppppppppppppppppppp

Mathematics
2 answers:
levacccp [35]2 years ago
6 0

Answer:

the answer is 97000

Step-by-step explanation:

miv72 [106K]2 years ago
5 0

Answer:

97000

Step-by-step explanation:

Just plug it in a calculator

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Tell whether the product SR is defined for S4×5 and R4×3 . If yes, give the product's dimensions as the following example: 4x3.
Rufina [12.5K]
No, the product is not valid. The number of columns of \mathbf S (five) should match the number of rows of \mathbf R (four).
8 0
2 years ago
What do you already know about linear relationships?
inna [77]

Answer:

A linear relationship (or linear association) is a statistical term used to describe a straight-line relationship between two variables. Linear relationships can be expressed either in a graphical format or as a mathematical equation of the form y = mx + b.

8 0
3 years ago
Write the exponent for the expression. <br> 1/8 x 1/8
kompoz [17]

Answer:

8^{-2}

Step-by-step explanation:

2^{-2} = 1/4

So:

8^{-2} = 1/8*1/8

5 0
2 years ago
A company tests the life of its rechargeable batteries over six months of use. They test how long a battery will last after a fu
jonny [76]

Answer:

18.6 months

Step-by-step explanation:

Given that :

Best fit line from scatterplot :

y=-12.05x +224.26

x = Number of month

y = charge on battery

Number of months a typical battery uses before being dead completely :

When battery is dead completely ; charge =0, y = 0

y = -12.05x + 224.26

0 = - 12.05x + 224.26

12.05x = 224.26

x = 224.26 / 12.05

x = 18.610788

Hence, 18.6 months before battery is completely dead.

4 0
2 years ago
Imagine that you need to compute e^0.4 but you have no calculator or other aid to enable you to compute it exactly, only paper a
labwork [276]

Answer:

0.0032

Step-by-step explanation:

We need to compute e^{0.4} by the help of third-degree Taylor polynomial that is expanded around at x = 0.

Given :

e^{0.4} < e < 3

Therefore, the Taylor's Error Bound formula is given by :

$|\text{Error}| \leq \frac{M}{(N+1)!} |x-a|^{N+1}$   , where $M=|F^{N+1}(x)|$

         $\leq \frac{3}{(3+1)!} |-0.4|^4$

         $\leq \frac{3}{24} \times (0.4)^4$

         $\leq 0.0032$

Therefore, |Error| ≤ 0.0032

4 0
3 years ago
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