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dexar [7]
2 years ago
14

Estimate the product of 41.03 x 0.513.

Mathematics
1 answer:
alisha [4.7K]2 years ago
8 0

Answer

Roughly, it would be 20.5

The exact answer is 21.05

Step-by-step explanation:

To roughly get the estimate you round it to the nearest 1st decimal place in this case. So it'd be 41.0 x 0.5

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If a1=4 and an =2an-1 – 1 then find the value of a4.
musickatia [10]
1 = 4

a2 = -2(a1) - 1
a2 = -2(4) - 1
a2 = -9

a3 = -2(a2) - 1
a3 = -2(9) - 1
a3 = -19

Can you do a4 and a5 and answer?
7 0
2 years ago
2x+3/4(4x+16)=7 <br> I need help finding x.. I don’t know how to do these problems with fractions :(
KIM [24]

Answer:

x=-1

Step-by-step explanation:

2x+3/4(4x+16)=7

2x+(3x12)=7--->distributive property

2x+3x+12=7

5x+12=7--->addition property

5x=-5--->subtraction peroperty

x= -1---?division property

5 0
2 years ago
there’s 13 snakes 11 lizards 7 turtles 8 tortoises how many fewer turtles and tortoises than snakes and lizard
Anna [14]

Answer: 9 fewer

Step-by-step explanation:

Add 13 and 11 which gives you 24. Then add 8 and 7 which gives you 15. Then subtract 24 - 15 which gives you a difference of 9.

8 0
2 years ago
Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [
Darya [45]

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

3 0
3 years ago
How do I solve this?
Ivenika [448]

\overline{XS}\cong\overline{YT}\Rightarrow 3m+7=4.2m+5\ \ \ |-7\\\\3m=4.2m-2\ \ \ \ |-4.2m\\\\-1.2m=-2\ \ \ \ |:(-1.2)\\\\m=\dfrac{2}{1.2}\\\\m=\dfrac{20}{12}\\\\m=\dfrac{5}{3}\to m=1\dfrac{2}{3}


\overline{YS}\cong\overline{XT}\Rightarrow3\dfrac{1}{2}r+2=2r+5\ \ \ \ |-2\\\\3\dfrac{1}{2}r=2r+3\ \ \ \ |-2r\\\\1\dfrac{1}{2}r=3\ \ \ |\cdot2\\\\3r=6\ \ \ \ |:3\\\\r=2

7 0
3 years ago
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