Thomas claims: The school bus will either be on time or late. The probability that it will be on time is therefore 0.5

5 less than the product of 8 and a number

zheka24 [161]

N-the number

the product of 8 and the number: 8 × n = 8n

5 less than the 8n: <u>8n - 5</u>

4 0

3 years ago

Read 2 more answers

Please help me solve

Law Incorporation [45]

CL ≈ AE and HG≈ZR and ∠L≈∠E as both triangles are isosceles so CL=HL and AE=EZ
so HL≈AE≈ZR≈EZ so according to side angle side ΔCLH andΔ AZE both are congruent so side CH≈AZ

4 0

3 years ago

Dale drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 7 hours. When Dale drove h

Yuliya22 [10]

Use a system of equations:

For his trip on the way there:
D = 7*s

For his trip on the way home you have:
D = 5*(s+18)

use substitution to solve

8 0

3 years ago

A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha

Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
the radius $r=5cm$
Punch is being poured into the bowl
The height at which the punch is increasing in the bowl is $\frac{dh}{dt} = 1.5$
the exposed area is a circle, (since the bowl is a hemisphere)
the radius of this circle can be written as $'a'$

what is being asked is the rate of change of the exposed area when the height $h = 2 cm$
the rate of change of exposed area can be written as $\frac{dA}{dt}$ .
since the exposed area is changing with respect to the height of punch. We can use the chain rule: $\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}$

and since $A = \pi a^2$ the chain rule above can simplified to $\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}$ -- we can call this Eq(1)

Solution:

the area of the exposed circle is

$A =\pi a^2$

the rate of change of this area can be, (using chain rule)

$\frac{dA}{dt} = 2 \pi a \frac{da}{dt}$ we can call this Eq(2)

what we are really concerned about is how $a$ changes as the punch is being poured into the bowl i.e $\frac{da}{dh}$

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

$r = \frac{a^2 + h^2}{2h}$

and rearrage the formula so that a is the subject:

$a^2 = 2rh - h^2$

now we can derivate a with respect to h to get $\frac{da}{dh}$

$2a \frac{da}{dh} = 2r - 2h$

simplify

$\frac{da}{dh} = \frac{r-h}{a}$

we can put this in Eq(1) in place of $\frac{da}{dh}$

$\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}$

and since we know $\frac{dh}{dt} = 1.5$

$\frac{da}{dt} = \frac{(r-h)(1.5)}{a}$

and now we use substitute this $\frac{da}{dt}$ . in Eq(2)

$\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}$

simplify,

$\frac{dA}{dt} = 3 \pi (r-h)$

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

$r = 5cm$

$h = 2cm$ from the question

$\frac{dA}{dt} = 3 \pi (5-2)$

$\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s$

5 0

3 years ago

HELP PLEASE, Domain and range problem

pentagon [3]

Answer:

Range is number of copies produced and set of values is; 1 ≤ N ≤ 200

Domain; Cost of publishing book in dollars; set of values are; $710 ≤ N ≤ $2700

Step-by-step explanation:

Range is a set of all the possible output values in a function while domain is the set of all possible input values.

Now, the function is given as;

C = 10N + 700

Where;

C is the cost of publishing the book in dollars

N is the number of copies of books produced

Thus, the domain will be a set of N values while Range will be a set of C values.

We are told that the first printing can produce up to 200 copies of the book.

That means a maximum of 200 books and a minimum of 1.

Thus;

Range is; 1 ≤ N ≤ 200

Maximum possible cost of the 200 books is;

C = 10(200) + 700

C = $2700

Minimum cost which will be for 1 book will be;

C = 10(1) + 700

C = $710

Thus,domain is;

$710 ≤ N ≤ $2700

6 0

2 years ago