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lana66690 [7]
3 years ago
10

Help please this is due tomorrow and I need this done

Mathematics
1 answer:
sukhopar [10]3 years ago
5 0

The answer is G. as per absolute value.

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I need help with his problem with details.
inysia [295]

Answer:

The solution is y> -1

Step-by-step explanation:

y- 6 > -7

y> -7+6 (add 6 on both sides)

y> -1

The graph should the the first option from the right, with the dot above -1.

Since y is greater than -1, the dot should be on -1 and the arrow should be pointing towards the right. It shouldn't be a coloured dot because the sign is 'greater than' (>) not 'greater or equal to' (≥)

5 0
3 years ago
Need Help please answer
djverab [1.8K]
For the 2nd part of 2. just plug in what you have for G in your previous graph into the equation. This will give you H for all 5 columns . Like 3×8(-1+5)= h = 3× 32= 96 so H should equal 96 and so on as far as this function.

For number 3. The equation is given so just plug in your T for time which is 3 seconds, so...-16(3)^2+90(3) = H the height at 3 seconds. I'm doing it in my head but should be the height is 414. You should also say whether it's ft or inches etc because the teacher or yourself left that out of the equation which is also vital lol.
6 0
3 years ago
HELP PLEASE IM BEGGING YOU
andriy [413]

Answer:

75th term in the sequence is 227.

Step-by-step explanation:

75 term of A.P is

a_75= a+ (n-1)d

Given series is 5,8,11,14.

hence a= 5 d= 8-5 = 3 , a_75.

a_75 = 5+(75-1)3

= 5+ (74)3

= 5 +(222)

= 227

6 0
2 years ago
What are the steps to converting 720 seconds into hours?
Nataly [62]
This conversion<span> of </span>720 seconds<span> to </span>hours<span> has been calculated by multiplying </span>720 seconds<span> by 0.0002 and the result is 0.2 </span>hours<span>.</span>
3 0
3 years ago
Read 2 more answers
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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