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allochka39001 [22]

1 year ago

5

If g (x) = one-third x and 3 x (StartFraction x Over 3 EndFraction), which expression could be used to verify that (one-third x)

(3 x) is the inverse of One-third (3 x)?
one-third (one-third x)

2 answers:

DENIUS [597]1 year ago

5 0

Answer:

c, 1/3(3x)

Step-by-step explanation:

Svet_ta [14]1 year ago

4 0

Answer:

c, 1/3(3x)

Step-by-step explanation:

You might be interested in

The area of an extra large circular pizza from Gambino's Pizzeria is 484\pi \text{ cm}^2484π cm 2 484, pi, space, c, m, start su

oksano4ka [1.4K]

Answer:

The diameter of an extra large pizza from Gambino's Pizzeria is $44\ cm$

Step-by-step explanation:

we know that

The area of a circle (circular pizza ) is equal to

$A=\pi r^{2}$

we have

$A=484\pi\ cm^{2}$

substitute in the formula and solve for the radius r

$484\pi=\pi r^{2}$

Simplify

$484=r^{2}$

$r=22\ cm$

Find the diameter

Remember that the diameter is two times the radius

$D=2(22)=44\ cm$

3 0

2 years ago

Read 2 more answers

1) Proiectiile catetelor unui triunghi dreptunghic pe ipotenuza au lungimile 9 cm si 25 cm. Aflati lungimea inaltimii din varful

Flauer [41]

Answer:

1) 15cm

2) left projection/h = h/right projection

Step-by-step explanation:

Question:

1) The projections of the legs of a right triangle on the hypotenuse have lengths of 9 cm and 25 cm. Find the length of the height at the top of the right angle.

2) In a right triangle the length of the hypotenuse is 34 cm, and the lengths of the projections of the legs on the hypotenuse are directly proportional to the numbers 0, (6) and 0.75. Calculate the length of the height corresponding to the hypotenuse.

Solution

1) The length of the height of a right angle triangle is also called the altitude.

Since there are no diagrams in the question, I sent a diagram of the right angle as an attachment to the solution.

The projections of the legs are 25cm and 9cm.

Hence, the longer projection length AD = 25cm and the shorter projection length DB = 9cm

In a right triangle, the altitude (height) drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the

geometric mean of these two segments (the two projections) and it's given by:

left projection/h = h/right projection

AD/h = h/DB

25/h = h/9

Cross multiply

h^2 = 25×9

h =√225 = 15cm

The length of the height at the top of the triangle = 15cm

2) Length of hypotenuse = 34

From the question, the lengths of the projections of the legs on the hypotenuse are directly proportional to the numbers 0, (6) and 0.75.

There is an error with the figures because the sum of the length of the projection of the legs should be equal to the hypotenuse but it isn't in this case.

To calculate the length of the height corresponding to the hypotenuse, we would use the same formula above.

left projection/h = h/right projection

To find each leg using question 1 above, each leg of the triangle is the mean proportional between the hypotenuse and the part of the hypotenuse directly below the leg.

Hypotenuse =34cm

Hyp/leg = leg/part

To find leg y, part for leg y = 25cm

34/y = y/25

y^2 = 34×25 = 850

y = √850 = 29.2cm

To find leg x, part for leg x = 9cm

34/y = y/9

y^2 = 34×9 = 306

y = √306 = 17.5cm

8 0

2 years ago

Sarah is planning to spend a week at her friend's summer house in Miami Beach. All meals will be provided by her friend's parent

Rus_ich [418]

Answer:

500$

Step-by-step explanation:

5 0

2 years ago

Triangles ABD and ACE are similar right triangles. which ratio best explains why the slope of AB is the same as the slope of AC

IgorC [24]

slope is rise over run.

Which segments from each triangle represent rise and which represent run?

little triangle: rise is BD and run is DA

big triangle: rise is CE and run is EA

Answer: D $\frac{BD}{DA} = \frac{CE}{EA}$

8 0

2 years ago

A survey found that 73% of adults have a landline at their residence (event A); 83% have a cell phone (event B). It is known tha

4vir4ik [10]

Answer:

3. What is the probability that an adult selected at random has both a landline and a cell phone?

A. 0.58

4. Given an adult has a cell phone, what is the probability he does not have a landline?

C. 0.3012

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that an adult has a landline at his residence.

B is the probability that an adult has a cell phone.

C is the probability that a mean is neither of those.

We have that:

$A = a + (A \cap B)$

In which a is the probability that an adult has a landline but not a cell phone and $A \cap B$ is the probability that an adult has both of these things.

By the same logic, we have that:

$B = b + (A \cap B)$

The sum of all the subsets is 1:

$a + b + (A \cap B) + C = 1$

2% of adults have neither a cell phone nor a landline.

This means that $C = 0.02$ .

73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)

So $A = 0.73, B = 0.83$ .

What is the probability that an adult selected at random has both a landline and a cell phone?

This is $A \cap B$ .

We have that $A = 0.73$ . So

$A = a + (A \cap B)$

$a = 0.73 - (A \cap B)$

By the same logic, we have that:

$b = 0.83 - (A \cap B)$ .

So

$a + b + (A \cap B) + C = 1$

$0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1$

$(A \cap B) = 0.75 + 0.83 - 1 = 0.58$

So the answer for question 3 is A.

4. Given an adult has a cell phone, what is the probability he does not have a landline?

83% of the adults have a cellphone.

We have that

$b = B - (A \cap B) = 0.83 - 0.58 = 0.25$

25% of those do not have a landline.

So $P = \frac{0.25}{0.83} = 0.3012$

The answer for question 4 is C.

4 0

2 years ago