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Irina18 [472]
3 years ago
11

What property is (x+y)+z=x+(y+z)

Mathematics
1 answer:
Agata [3.3K]3 years ago
4 0
Associative Property if Addition
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Peter threw the football 72 feet. How many yards did he throw the football ?
Darya [45]

1 yard = 3 feet.

Divide total feet by 3 to get yards:

72 feet / 3 = 24 yards.

3 0
3 years ago
Nico is saving money for his college education. He invests some money at 9% and 800 less than that amount at 6%. The investments
vampirchik [111]

Answer:

$1800 at 9% and $1000 at 6%

Step-by-step explanation:

Not needed

5 0
3 years ago
Can anyone help<br> Me with this problem plz
seropon [69]
First you need common denominator
x/2 - 6/2 = 34/2
now , eliminate denominator 
x - 6 = 34 add 6 to both sides and simplify
x=40
3 0
3 years ago
Which ordered pair is a reflection of (3,7) across the x-axis?
Alla [95]

Answer:

(3,-7)

Step-by-step explanation:

Using the algebraic rule for Rx-axis

(x,y)---->(x,-y)

(3,7)--->(3,-7)

5 0
2 years ago
Find the point P on the line y=2x that is closest to the point (20,0) .what is the least distance between pans (20,0)?
Vsevolod [243]
The closest point on a line, to another point, will be a point that's on a normal of that line, or a line that is perpendicular to it, notice picture below

so, y = 2x, has a slope of 2, a perpendicular line to it, will have a slope of negative reciprocal that, or \bf 2\qquad negative\implies -2\qquad reciprocal\implies \cfrac{1}{-2}\implies -\cfrac{1}{2}

so, we know that line passes through the point 20,0, and has a slope of -1/2

if we plug that in the point-slope form, we get \bf y-0=-\cfrac{1}{2}(x-20)\implies y=-\cfrac{1}{2}x+10

now, the point that's on 2x and is also on that perpendicular line, is the closest to 20,0 from 2x, thus, is where both graphs intersect, as you can see in the graph

thus  \bf 2x=-\cfrac{1}{2}x+10  solve for "x'

------------------------------------------------------------

not sure on the 2nd part, but sounds like, what's the distance from that point to 20,0, well, if that's the case, just use the distance equation

\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;&({{ \square }}\quad ,&{{ \square }})\quad &#10;%  (c,d)&#10;&({{ \square }}\quad ,&{{ \square }})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}


6 0
3 years ago
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