Please help thank you

Someone help me with this please

disa [49]

Answer:

5

Step-by-step explanation:

with Proportionality

6 0

1 year ago

~khan academy 7th grade

pickupchik [31]

Question:

Which of the following expressions represents the distance between 1.7 and −1/2 on a number line?

Choose 1 answer:

$A.\ |1.7 - 1/2|$

$B.\ |1.7 -(-1/2)|$

C. None of the above

Answer:

$B.\ |1.7 -(-1/2)|$

Step-by-step explanation:

Given

Numbers: 1.7 and -1/2

Required

Determine the distance

To calculate the distance, we start by calculating the difference between the given numbers;

$Difference = 1.7 - (-1/2)$

Then, we calculate the distance;

<em>In this case, the distance is meant to be positive; So, we'll introduce the symbol for absolute value;</em>

Hence,

$Distance = |Difference|$

Substitute $1.7 - (-1/2)$ for Difference;

$Distance = |1.7 - (-1/2)|$

<em>Hence, the correct option is B</em>

4 0

2 years ago

Read 2 more answers

Let f and g be the functions defined by f(x) = 10^ (x+2 / 3) and g(x) = log (x3 / 100) for all positive real numbers,

Vaselesa [24]

Answer:

F(x) and g(x) are not inverse functions.

Step-by-step explanation:

In order to calculate the inverse function of a function, we have to isolate X and after that, we change the variables.

As our function f(x) is a exponentian function, we can apply logarithm with base 10 (log) in both sides in order to isolate X. Remember that log10=1.

$[tex]y=10^{(x+\frac{2}{3}) }\\\\log y=log 10^{(x+\frac{2}{3}) }\\log y = (x+\frac{2}{3}) . log10\\\frac{log Y}{log10} = (x+\frac{2}{3})\\\frac{log Y}{1} = (x+\frac{2}{3})\\log Y-\frac{2}{3}=x$ [/tex]

Now we change the variables.

$F(x)=log x-\frac{2}{3}$

F(x) and g(x) are not inverse functions.

4 0

2 years ago

What is 79.34 to one decimal point

IceJOKER [234]

I’m assuming you mean to round to the nearest tenth? If so, it’s 79.3. If that’s not what you’re asking i don’t know what else you mean

5 0

2 years ago

f the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced, w

LUCKY_DIMON [66]

The frictional force between the tires and the road prevent the car from skidding off the road due to centripetal force.

If the frictional force is less than the centripetal force, the car will skid when it navigates a circular path.

The diagram below shows that when the car travels at tangential velocity, v, on a circular path with radius, r, the centripetal acceleration of v²/ r acts toward the center of the circle.

The resultant centripetal force is (mv²)/r, which should be balanced by the frictional force of μmg, where μ = coefficient of kinetic friction., and mg is the normal reaction on a car with mass, m.

This principle is applied on racing tracks, where the road is inclined away from the circle to give the car an extra restoring force to overcome the centripetal force.

8 0

2 years ago