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USPshnik [31]
3 years ago
12

5. WHAT IS THE SALE PRICE OF THIS BAG IF IT IS MARKED PHP2735 WITH 33% DISCOUNT?​

Mathematics
1 answer:
Alex777 [14]3 years ago
8 0

Answer:

1832.45

Step-by-step explanation:

2735 - 33% = 1832.45

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What is 2a +16 what does that =
kirill115 [55]
A will equal 8. can i get a brainliest plz
7 0
3 years ago
Match each object with the characteristics it represent
Aleks [24]

We need the object and the characteristics.

6 0
3 years ago
What is the midpoint of the line segment with endpoints (-5.5, -6.1) and (-0.5, 9.1)?
Margarita [4]

Answer:

(-5/2, 3/2)

Step-by-step explanation:

[(-5.5+.5)/2, (9.1-6.1)/2)} = (-5/2, 3/2)

3 0
3 years ago
One zero of x^3 - 4x = 0 is 0 what are the other zeros of the function
myrzilka [38]
So you have x^3 - 4x = 0. What you can do is pull out an x from both x^3 and - 4x so it looks like this:

x( {x}^{2} - 4) = 0

Then you can find a number that makes the part inside the parentheses turn into zero. For beginners, it may be easier to write it out seperately and solve for x.

{x}^{2} - 4 = 0

We need to solve for x, so the first step is to add 4 to both sides, so we get something like this:

{x}^{2} = 4

Then, we can square root both sides to get rid of the power on the x, so it looks like this:

x = \sqrt{4}

Now, every square root has two answers, a positive and a negative. If we look at the bottom example:

{2}^{2} = 4

{( - 2)}^{2} = 4

We can see that both -2 and 2 to the power of two will equal to 4.

So finally, we get:

x = - 2 \: and \: 2

These are the other 'Zero's for the original function. If you are not sure of what a 'Zero' is, it is where the function crosses over the x-axis on a graph.
5 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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