Answer:
=25343 grams or 25 kg 343 g.
Step-by-step explanation:
1 kg = 1000 g
∴ 4 kg = 4000 grams
and the 4 kg 250 g would equal = 4250 grams.
∴ 2 kg = 1000 g
2 kg 90 g = 2000 grams + 90 g = 2090 grams
∴ 19 kg = 19000 grams
and 19 kg 3 grams = 19000 grams + 3 grams = 19003 grams
now we will add all of the values together to get the final answer.
4250 grams + 2090 grams + 19003 grams = 25343 grams or 25 kg 343 g.
Hope this helps!!
80,000.
To round up to the nearest ten thousand, we would see if the next lower place value has a amount either 5 or higher or 4 or lower.
If it’s five or higher, we round up.
If it’s four or lower, we round down.
We can see that the number in the ten thousands place is 8. The number in the place value after 8 is 2.
2<5 so we round down.
The answer is 80,000.
Hope this helps!
Answer:

Step-by-step explanation:
Let
x ----> amount that Jethro owes in taxes
we know that
Amount that he made last year 
His tax rate 
Applying proportion
solve for x

I am sorry I can help more but
Google can graph lines if that helps
You just look up graph of a line with your numbers
Answer:
Part 1)
Bob's mistake was to have used the cosine instead of the sine
The measure of the missing angle is 
Part 2) The surface area of the pyramid is 
Step-by-step explanation:
Part 1)
Let
x----> the missing angle
we know that
In the right triangle o the figure
The sine of angle x is equal to divide the opposite side angle x to the hypotenuse of the right triangle


Bob's mistake was to have used the cosine instead of the sine
Part 2) we know that
The surface area of the square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces
so
![SA=b^{2}+4[\frac{1}{2}(b)(h)]](https://tex.z-dn.net/?f=SA%3Db%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28h%29%5D)
where
b is the length side of the square
h is the height of the triangular lateral face
In this problem
-------> by an 45° angle
so



Find the value of b

Find the surface area
![SA=12^{2}+4[\frac{1}{2}(12)(6)]=288\ cm^{2}](https://tex.z-dn.net/?f=SA%3D12%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%2812%29%286%29%5D%3D288%5C%20cm%5E%7B2%7D)