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tia_tia [17]
3 years ago
14

£108 in the ratio 7:5

Mathematics
1 answer:
dybincka [34]3 years ago
8 0

Answer:

45:63

Step-by-step explanation:

7+5=12

12u-->108

1u--> 9

5u-->45

7u-->63

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Estimate the value of 9.9 power of 2 x 1.79 <br> pls some one help me
dybincka [34]

Answer:

The correct answer is 175.4379

Step-by-step explanation:

(9.9)^2 • 1.79

=98.01 • 1.79

=175.4379

3 0
4 years ago
When Ms.Sugar turned on her oven, the temperature inside was 70 degrees. The temperature began to rise at a rate of 20 degrees p
zavuch27 [327]
It took 14 minutes
70 degrees +(14minutes times 20degrees) = 350 degrees.
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3 0
4 years ago
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How do you graph f(x)=-3x^2+5
UNO [17]

Answer:

Make sure you plot your points in a table and set it up on a graph

Step-by-step explanation:

3 0
4 years ago
Pls help
solong [7]
Ok let’s solve it

5(x-2)^2-20=0

first let’s foil (x-2)

5(x^2-4x+4) -20=0

now distribute the 5

5x^2 -20x +20 -20 = 0

combine like terms

5x^2-20x=0

take the gcf

5x(x-4)=0

x=0, 4

solutions are (4,0) and (2, -20) because the original vertex form a(x-h)^2+k

7 0
3 years ago
How long will it take for $2500 to double if it is invested at 6% annual interest compounded 6 times a year? Enter exact calcula
12345 [234]

Answer:

Part 1) t=11.610\ years

Part 2) t=11.552\ years

Step-by-step explanation:

Part 1) we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

A=\$5,000\\P=\$2,500\\ r=6\%=6/100=0.06\\n=6  

substitute in the formula above

5,000=2,500(1+\frac{0.06}{6})^{6t}  

2=(1.01)^{6t}  

Apply log both sides

log(2)=log[(1.01)^{6t}]

log(2)=(6t)log(1.01)  

solve for t

t=log(2)/[6log(1.01)]  

t=11.610\ years

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

A=P(e)^{rt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest in decimal  

t is Number of Time Periods  

e is the mathematical constant number

we have  

A=\$5,000\\P=\$2,500\\ r=6\%=6/100=0.06  

substitute in the formula above

5,000=2,500(e)^{0.06t}

2=(e)^{0.06t}  

Apply ln both sides

ln(2)=ln[(e)^{0.06t}]

ln(2)=(0.06t)ln(e)

ln(2)=(0.06t)

t=ln(2)/(0.06)

t=11.552\ years

3 0
3 years ago
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