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3 years ago

What are the truth values of Q v ~R?

Mathematics

1 answer:

GREYUIT [131]3 years ago

4 0

Answer:

ttft

Step-by-step explanation:

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Petrov ran 7 laps around a track.Each lap around the track is a distance of 1/4 mile. What is the total distance Petrov ran in f

Alla [95]

Answer:

9240 feet

Step-by-step explanation:

Petrov ran 7 laps around a track.Each lap around the track is a distance of 1/4 mile.

Step 1

The total distance Petrov ran in miles is calculated as:

1 lap = 1/4 mile

7 laps = x mile

Cross Multiply

x = 7 × 1/4 mile

x = 1.75 miles

Step 2

What is the total distance Petrov ran in feet?

1 mile = 5280 feet

1.75 miles = x

Cross Multiply

x = 1.75 × 5280 feet

x = 9240 feet

Therefore, the total distance Petrov ran in feet is 9240 feet

3 0

3 years ago

In ΔHIJ, the measure of ∠J=90°, HI = 8 feet, and JH = 1.9 feet. Find the measure of ∠H to the nearest tenth of a degree.

artcher [175]

Answer:

21.9°

Step-by-step explanation:

In ΔHIJ, the measure of ∠J=90°, HI = 8 feet, and JH = 1.9 feet. Find the measure of ∠H to the nearest tenth of a degree.

We solve this above question using the Sine rule

a/sin A = b/sin B

In ΔHIJ, the measure of ∠J=90°, HI = 8 feet, and JH = 1.9 feet.

Hence:

HI/∠J = JH/∠H

= 8/sin 90° = 1.9/sin ∠H

Cross Multiply

∠H = arc sin(sin 1.9 × 90/8)

∠H = 21.9°

5 0

2 years ago

What is the range of this data (3,3,08,7,10,2,6,12,0) A 7.5 B 6.5 C 10 D 12

solniwko [45]

The range is 12. (12-0)

6 0

3 years ago

A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y

siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32

Given that 100 beetles are released into a field.

So that means initial number of beetles P=100

Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

$A=P(1+r)^n$ where A is number of beetles at any year n.

Plug the given values into above formula we get:

$A=100(1+0.32)^n$

$A=100(1.32)^n$

now plug n=10 years

$A=100(1.32)^{10}=100(16.0597696605)=1605.97696605$

Hence answer is approx 1606 beetles will be there after 20 years.

Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

$A=100(1.32)^{20}=100(257.916201549)=25791.6201549$

Hence answer is approx 25791 beetles will be there after 20 years.

Now we have to find time for 100000 beetles so plug A=100000

$A=100(1.32)^n$

$100000=100(1.32)^n$

$1000=(1.32)^n$

$log(1000)=n*log(1.32)$

$\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n$

33.174666862=n

Hence answer is approx 33 years.

4 0

3 years ago

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PLEASE HELP ASAP

Nikitich [7]

10 x 10 x 10 = $10^{3}$

4 0

3 years ago

Read 2 more answers