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SCORPION-xisa [38]
3 years ago
12

an online newspaper had 350,080 visitors in October 350,489 visitors in November and 305,939 visitors in December what is the or

der of the months from greatest to least number of visitors
Mathematics
1 answer:
bekas [8.4K]3 years ago
4 0
November, October, December
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X^2=14x-49. i just dont know how to do the steps to solve this equation
KonstantinChe [14]
x^2=14x-49 \\ \\ x^2 - 14x + 49 = 0 \ / \ move \ terms \\ \\ x^2 - 2(x)(7) + 7^2 = 0 \ / \ rewrite \\ \\ (x - 7)^2 = 0 \ / \ square \ of \ difference \ method \\ \\ x - 7 = 0 \ / \ square \ root \ each \ side \\ \\ x = 7 \ / \ add \ 7 \\ \\

The answer is: x = 7.
5 0
3 years ago
Which system of equations can be used to find the roots of the equation 4x^5-12x^4+6x=5x^3-2x?
Sveta_85 [38]
^{(1)}\ \ 4x^5-12x^4+6x=5x^3-2x\ \ ^{(2)}\\\\  \left\{\begin{array}{ccc}y=4x^5-12x^4+6x\ \ \ ^{(1)}\\y=5x^3-2x\ \ \ ^{(2)}\end{array}\right


Answer: system nr 4.
7 0
3 years ago
Rewrite this expression in addition<br> -8-(-12)-7
Andrej [43]

Answer:

-8+(+12)+)

Step-by-step explanation:

dont know

4 0
3 years ago
Please answer! Explain all steps!
MatroZZZ [7]
We will be using the formulas:
speed=distance/time
time=distance/speed
distance=speed×time
First let's find out Diane's rate of swimming. We can measure this by finding the slope (y/x) of a given coordinate on the graph. One point is (10,15), so you do 15/10=1.5m/s
Now for Rick's rate of swimming, just take a pair of values from the table. 12.5/10=1.25m/s
By the way m/s is metres per second for this
So at a constant speed of 1.5m/s, Diane swam 150m in 150/1.5= 100 seconds, or 1 minute 40 seconds
And at a constant speed of 1.25m/s, Rick swam 150m in 150/1.25= 120 seconds, or 2 minutes.
So the difference between their two times is 20 seconds
7 0
3 years ago
Read 2 more answers
URGENT HELP NEED!!! I WILL REWARD LOTS OF POINTS FOR THIS QUESTION!!! ANY IRRALEVENT OR SILLY QUESTION WILL BE INSTANTLY REPORTE
Ipatiy [6.2K]

Let the points be A,B,C

  • A(0,0)
  • B(24,16)
  • C(8,20)

#A

\\ \bull\sf\leadsto AB=\sqrt{(24-0)^2+(16-0)^2}=\sqrt{24^2-16^2}=\sqrt{576-256}=\sqrt{320}=17.4units(leg3)

\\ \bull\sf\leadsto AC=\sqrt{(8-0)^2+(20-0)^2}=\sqrt{8^2+20^2}=\sqrt{64+400}=\sqrt{464}=21.4units(leg1)

\\ \bull\sf\leadsto BC=\sqrt{(24-8)^2+(26-20)^2}=\sqrt{16^2+(-4)^2}=\sqrt{256+16}=\sqrt{272}=16.4units(leg2)

#B

We have to find perimeter

\\ \bull\sf\leadsto Perimeter=AB+AC+BC

\\ \bull\sf\leadsto Periemter=17.4+16.4+21.4

\\ \bull\sf\leadsto Perimeter=55.2units

#C

The length of leg3=17.4units

5 0
2 years ago
Read 2 more answers
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