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3 years ago

11

Some help at 1 and 2 pleaseeeeeeeeee I really need help

2 answers:

Vaselesa [24]3 years ago

7 0

The first one would be 12+14+18+12+15. She biked about 71 miles.

I did this by rounding the decimal places. 5 and up = round up. 4 or less= stays the same. I'll answer #2 in a second, I have to do something :)

Jlenok [28]3 years ago

4 0

She biked about 5.9 more miles on wednesday

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Enter the decimal equivalent of 9/12

AnnyKZ [126]

Answer:

.75, I hope that this has helped

5 0

3 years ago

Complete the square to solve the equation below.

sasho [114]

Hello there!

x² + x = 7/4
x² + - 7/4 = 0

Now we gonna use the quadratic formula to find x
a= 1
b=1
c = -1.75

x = -b+/-√b² -4ac all of them divide by 2a
x = -(1)+/-√(1)² - (4)(1)(-1.75) all of them divide by 2(1)
x= -1+/-√8 all of them divide by 2
x = -1/2 + √2 or x = -1/2 - √2

The correct option is option C

I hope that helps!

4 0

3 years ago

6. Find average of the following<br> expressions (4-2x), (-7-3x), and<br> (11x+6)

svetoff [14.1K]

Answer:

2x+1

Step-by-step explanation:

(4-2x), (-7-3x),(11x+6)

Add the three expressions

(4-2x)+ (-7-3x)+(11x+6)

Combine like terms

-2x-3x+11x+4-7+6

6x+3

Divide by the number of expressions which was 3

(6x+3)/3

2x+1

The average is 2x+1

7 0

3 years ago

Read 2 more answers

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

So

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago

Line B will be graphed on the same grid. The only solution to the system of linear equations formed by Line A and Line B will oc

Nady [450]

Answer:

Step-by-step explanation:

7 0

2 years ago