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natali 33 [55]
3 years ago
11

3.5^1.6 round answer to nearest thousandth​

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
5 0

The answer would be 7.42

Use desmos online calculator

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PLEASE HELP! WILL UPVOTE!
bagirrra123 [75]
Just "plug n chug" as my teacher would say. In other words, replace the letters with the values. 

Choice A is (7,19) which means x = 7 and y = 19. Let's replace the x and y value with these numbers in both equations. If we get a true equation, then we found the answer. 

2x-y = -5
2*7-19 = -5
14-19 = -5
-5 = -5 ... true
So that works. Let's try the other equation
x+3y = 22
7+3*19 = 22
7+57 = 22
64 = 22 ... false 
So choice A is NOT the answer because of this. Let's try out choice B
2x-y = -5
2*4-9 = -5
-1 = -5 ... false
so B doesn't work either. How about C? 
2x-y = -5
2*8-21 = -5
-5 = -5 ... that works
x+3y = 22
8+3*21 = 22
71 = 22 ... but this is false
So D has to be the answer. Let's check to confirm or not
2x-y = -5
2*1-7 = -5
-5 = -5 ... true
x+3y = 22
1+3*7 = 22
22 = 22 ... true
BOTH equations are true at the same time so (1,7) is definitely the final answer


6 0
3 years ago
You build and sell toy boxes. So far, you have built 15 toy boxes that you will sell for $60 each. How much money will you have
nignag [31]

15 x $60 = 900

20 x $60 = 1200

900 + 1200 = 2100

$2100 - answer

6 0
3 years ago
(3x + 10)<br> (5x-10)<br> A)x=30 B.)*= 28<br> c)x=25 Dx=10
MrMuchimi

Answer:D. x=10

Step-by-step explanation:

4 0
3 years ago
Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located
arsen [322]
Check the picture below.

\bf \textit{using the pythagorean theorem}&#10;\\\\&#10;c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\&#10;-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}&#10;\\\\\\&#10;csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

3 0
3 years ago
I needdd helppppppppp real bad!!!!! Plwesss help!!! TYSMMMM!!
BaLLatris [955]
Soz man i use the metric system
6 0
3 years ago
Read 2 more answers
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