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dimulka [17.4K]
3 years ago
13

What percent is 72 of 120???

Mathematics
2 answers:
just olya [345]3 years ago
8 0
Well, first you set up a proportion. x/100 = 72/120 (remember is over of) and you cross multiply. 100 x 72= 7200. 120x=7200. Then you divide both sides by 120, and you get 60%
ollegr [7]3 years ago
4 0
What percent is 72 of 120???
120*x\% = 72 \\  \\ \\  120*  \dfrac{x}{100} = 72 \\  \\  \\  \dfrac{x}{100} = \dfrac{72}{120}  \\  \\  \\ x= \dfrac{72}{120} * 100\qquad \to  \boxed{x= 60\% }

60% *120 = 72
0,6 *120   = 72 
       72 = 72   


I wish you much success in your studies!!!!
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Answer:

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Step-by-step explanation:

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7 0
3 years ago
Read 2 more answers
What is the value of -3|15 - s| + 2s3 when's = -3
slamgirl [31]
<span>-3|15 - s| + 2s^3 when s = -3 is
</span>-3|15 - (-3)| + 2(-3)^3 = <span>-3|15 + 3| + 2(-27) = </span><span>-3|18| + (-54) = -3(18) - 54 = -54 - 54 = -108</span>
6 0
3 years ago
I need help doings this​
julia-pushkina [17]

Answer:

Either A or C, I'm not quite sure tho

Step-by-step explanation:

The line is on the negative side of the graph

6 0
2 years ago
6. An ostrich cannot fly, but it is able to run fast. Suppose an ostrich runs east for 7.95 s and then runs 161 m south, so that
Liula [17]

Answer:

159 m

Step-by-step explanation:

From the information given:

It was stated that if the ostrich ran towards the east direction in 7.95 s, let say the distance from the starting point is O towards the east side E, let called the distance towards the east side to be OE.

Again, the ostrich then runs in the south direction for 161 m, let the distance be OS.

Also, let the magnitude of the resultant displacement between the east direction to the south direction be ES = 226m.

We are to find, the magnitude of the ostrich's eastward component.

i.e. The distance traveled from the center to the east direction within the time frame of 7.95 s.

Using the Pythagoras rule:

ES² = OE² + OS²

226² = OE² + 161²

226² -  161² = OE²

OE²  =  226² -  161²

OE²  = 51076 - 25921

OE²  =  51076 - 25921

OE²  =  25155

OE = \sqrt{25155}

OE =  158.60 m

OE ≅ 159 m

Thus, the magnitude of the ostrich's towards the eastward component. = 159 m.

4 0
3 years ago
How do I solve 1 and 2?
LuckyWell [14K]
H=+15 m
v=+5 m/s
Ball hits the ground when h(t)=-15m
h(t)=-9.8t^2+vt+h
=>
-15=-9.8t^2+5t+15
9.8t^2-5t-30=0
Solve for t, using quadratic formula,
t=-1.513 or t=2.023
reject negative root due to context, so
t=2.023 seconds

2)
h(t)=-16t^2+20t+8
a. height before pitch is when t=0, or h(0)=8
b. highest point reached when h'(t)=-32t+20=0 => t=5/8 seconds
c. highest point is t(5/8)=-16(5/8)^2+20(5/8)+8=47/5=9.4 m
d. ball hits ground when h(t)=0 => solve t for h(t)=0
=> t=-0.3187 seconds or t=1.569 seconds.
Reject negative root to give
time to hit ground = 1.569 since ball was pitched.

5 0
3 years ago
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