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olchik [2.2K]
2 years ago
6

Simplify the expression 1+2*[(3+1)*5+3]

Mathematics
2 answers:
Lady bird [3.3K]2 years ago
7 0

1 + 2 *[(3+1)*5+3]

=  1 + 2*[(4)*5+3]

=  1 + 2*[20+3]

=  1 + 2 * 23

= 1 + 46

= 47

Answer

c. 47

kondaur [170]2 years ago
6 0
C.47 is the answer to the expression
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Which statement is true about the given information ?
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3 years ago
Which number is rational? 2 square root, pi ,10 square root ,16 squeare root
ki77a [65]

Answer:

Square root of 16 is rational beacuse Square root of 16=4 Whlie the others are Irational.

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
PLEASE HELP!!
PSYCHO15rus [73]

<u><em>Answer:</em></u>

The bird is approximately 9 ft high up in the tree

<u><em>Explanation:</em></u>

The required diagram is shown in the attached image

Note that the tree, the cat and the ground form a right-angled triangle

<u>Therefore, we can apply special trigonometric functions</u>

<u>These functions are as follows:</u>

sin(\alpha)=\frac{opposite}{hypotenuse} \\ \\ cos(\alpha)=\frac{adjacent}{hypotenuse} \\ \\tan(\alpha)=\frac{opposite}{adjacent}

<u>Now, taking a look at our diagram, we can note the following:</u>

α = 25°

The opposite side is the required height (x)

The adjacent side is the distance between the cat and the tree = 20 ft

Therefore, we can use the <u>tan function</u>

<u>This is done as follows:</u>

tan(\alpha)=\frac{opposite}{adjacent}\\ \\ tan(25)=\frac{x}{20}\\ \\x=20tan(25) = 9. 32 ft which is 9 ft approximated to the nearest ft

Hope this helps :)

5 0
3 years ago
In a set of ordered pairs, the y values repeated while the x values did not repeat.
LiRa [457]
Answer: D

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7 0
3 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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