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igor_vitrenko [27]
3 years ago
15

1.If 70% is £14, what is the full cost

Mathematics
1 answer:
Artist 52 [7]3 years ago
5 0
1) £20
2) 18/4 * 7 = 31,5msquared
3) X= (3-1)/5 = 2/5 = 0.4
4) A(n) = 2 + 5n
5) t(3) = 6*3 + 2 = 20
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Which equation will be parallel to y=6 A)Y= -6x B)X=6 C)y=1 D)Y=6x
galben [10]

Answer:

A

Step-by-step explanation:

4 0
3 years ago
If it’s right I will mark you Brainly!!
Free_Kalibri [48]

Answer:

y=x+20

or

y= 1x + 20

Step-by-step explanation:

The formula for slope is y= mx+ b

m is the slope and b is the y-intercept (where the line touches the y-axis

4 0
3 years ago
The level of CO2 emissions, f(x), in metric tons, from the town of Fairfax x years after they started recording is shown in the
Temka [501]
1) The average increase in the level of CO2 emissions per year from years 2 to 4 is:
Average=[f(4)-f(2)]/(4-2)=(29,172.15-26,460)/2=2,712.15/2=1,356.075 metric tons. The first is false.

2) The average increase in the level of CO2 emissions per year from years 6 to 8 is:
Average=[f(8)-f(6)]/(8-6)=(35,458.93-32,162.29)/2=3,296.64/2=1,648.32 metric tons. The second is false.

3) The average increase in the level of CO2 emissions per year from years 4 to 6 is:
Average=[f(6)-f(4)]/(6-4)=(32,162.29-29,172.15)/2=2,990.14/2=1,495.07 metric tons. The third is false.

4) The average increase in the level of CO2 emissions per year from years 8 to 10 is:
Average=[f(10)-f(8)]/(10-8)=(39,093.47-35,458.93)/2=3,634.54/2=1,817.27 metric tons. The fourth is true.

Answer: Fourth option: The average increase in the level of CO2 emissions per year from years 8 to 10 is 1,817.27 metric tons.
7 0
3 years ago
(20×2)20+2)20×2)+20(+41+2=​
andrew-mc [135]

Answer:

123

Step-by-step explanation:

BODMAS

1.multiply 20 by 2

2.20 ×2

3. 40+22+40+20+43=123

3 0
3 years ago
What percent of $900 equals 810?? Show what you did!!
KIM [24]

Answer:

x% of $900=$810

x/100×$900=$810

x=$810/9=90% required percentage is 90%.

6 0
3 years ago
Read 2 more answers
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