Label each pointon the number line with the correct value

A publisher needs to send many books to a local book retailer and will send the books in a combination of small and large boxes.

mote1985 [20]

Answer:

5 large boxes

3 small boxes

Step-by-step explanation:

Create two equations to represent the problem.

let "a" be the number of small boxes

let "b" be the number of large boxes

20a + 40b = 260 This equation shows the numbers of books

a + b = 8 This equations shows the number of boxes

Solve the system of equations (solve for a and b). We can solve using the substitution method.

Rearrange a + b = 8 to isolate one of the variables.

a = 8 - b New equation that represents "a"

Since we know an equation for "a", we can substitute what "a" equals into the other equation. There will only be one variable in the equation, so we can solve by isolating. Isolate by doing reverse operations.

Substitute "a" for 8 - b

20a + 40b = 260

20(8 - b) + 40b = 260 Distribute 20 over the brackets by multiplying

160 - 20b + 40b = 260 Collect like terms (numbers with same variables)

160 + 20b = 260 Start isolating "b". Subtract 160 from both sides

20b = 260 - 160

20b = 100 Divide both sides by 20

b = 5 Number of large boxes

Substitute "b" for 5 in the simplest equation

a + b = 8

a + 5 = 8 Subtract 5 from both sides to isolate "a"

a = 8 - 5

a = 3 Number of small boxes

Therefore there were 5 large boxes and 3 small boxes sent.

3 0

3 years ago

I’m confused on this one

erastovalidia [21]

In the chart of the triangle it shows that the variable b is the base. It also give us the equation for area 1/2bh (half of base * height)

Now we just plug in the values we are given

1/2bh

1/2(10*13)

1/2(130)

65

The area of the triangle is 65 ft

4 0

3 years ago

Read 2 more answers

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

Please help with my geometry

faust18 [17]

Answer:

www.mahtpapa.com

Step-by-step explanation:

5 0

3 years ago

Simplify.

natima [27]

Its 2y+7z hope dis helps

5 0

3 years ago

Read 2 more answers