All categories

2 years ago

How do i estimate .48

Mathematics

1 answer:

cricket20 [7]2 years ago

3 0

To estimate, we can round to the nearest tenth.

How to round:

5 and above, round up.

4 and below, round down.

0.48 -> 0.5

Best of Luck!

You might be interested in

Ellis is painting wooden fenceposts before putti by then in his yard. They are each 5 feet tall and have a diameter of 1 foot. T

lawyer [7]

Answer:

39.25

Step-by-step explanation:

diameter. is 1 therefore radius is 1/2

formula is πr²h

where

h is height=5

=3.14x½²x5

=3.925 for one paint

therefore for 10 paints, we have

=3.925x10

=39.25

3 0

2 years ago

Will get brainliest!!!!!!!!!!! 1 − 3(3 + 6b) = −6(3b + 1) − 2

Arlecino [84]

Answer:

b = infinity

Step-by-step explanation:

1 − 3(3 + 6b) = −6(3b + 1) − 2

1 - 9 - 18b = -18b + 1 -2

1 - 9 = 1 - 2

-8 = -1

7 0

2 years ago

A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times.

tatyana61 [14]

Answer:

(a) $g(n)=0.05\cdot 2^n$

(b) $g^{-1}(n)=\log_{2}20g(n)$

Step-by-step explanation:

Part A

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

$g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n$

Part B

$g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)$

Part C

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

$g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43$