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3 years ago

8

Jill's boat holds 42 containers.Each container can hold 8 big fish. How many big fish can Jill put in her 42 containers?

1 answer:

jek_recluse [69]3 years ago

6 0

The answer is 336 hope I helped

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What is the actual height of the building shown in the model?

julsineya [31]

Answer:

The actual height is 150ft

Step-by-step explanation:

Given

$Width : Height = 2cm : 5cm$

$Actual\ width = 60ft$

Required

Determine the actual height

Substitute 60ft for width in: $Width : Height = 2cm : 5cm$

$60ft:Height =2cm:5cm$

Express as fraction

$\frac{60ft}{Height} =\frac{2cm}{5cm}$

$\frac{60ft}{Height} =\frac{2}{5}$

Make height the subject

$Height = \frac{5}{2} * 60ft$

$Height = 5 * 30ft$

$Height = 150ft$

7 0

2 years ago

Read 2 more answers

Write the area A of a square as a function of its perimeter P. ...?

Svetradugi [14.3K]

The answer is A = P²/16

The perimeter P of a square is sum of its sides s: P = s + s + s + s = 4s
The area A of a square with side s is: A = s * s = s²

Step 1: Solve s from the formula for the perimeter.
Step 2: substitute s from the formula for the perimeter into the formula for the area.

Step 1:
P = 4s
s = P/4

Step 2:
A = s²
s = P/4
A = (P/4)²
A = P²/4²
A = P²/16

4 0

3 years ago

Find lim h→0 f(2+h)-f(2)/h if f(x)=x^2+x+1

bekas [8.4K]

Answer:

5

Step-by-step explanation:

$\displaystyle \lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$

$\displaystyle\lim_{h \to 0} \frac{(2+h)^2 + 2 + h + 1 - 2^2 - 2 - 1}{h}$

$\displaystyle\lim_{h \to 0} \frac{4 + 4h + h^2 + h - 4}{h}$

$\displaystyle\lim_{h \to 0} \frac{h^2 + 5h}{h}$

$\displaystyle\lim_{h \to 0} h + 5$

= 5

8 0

2 years ago

Please Help!!!! Find two consecutive odd numbers whose product is 47 more than their sum.

fenix001 [56]

7 and 9

7 x 9 = 63

7+9=16

63-16 = 47

7 0

2 years ago

The power generated by an electrical circuit (in watts) as function of its current x (in amperes) is modeled by: P(x)=-12x^2 +12

Minchanka [31]

Given:

The power generated by an electrical circuit (in watts) as function of its current x (in amperes) is modeled by:

$P(x)=-12x^2+120x$

To find:

The current that will produce the maximum power.

Solution:

We have,

$P(x)=-12x^2+120x$

Here, leading coefficient is negative. So, it is a downward parabola.

Vertex of a downward parabola is the point of maxima.

If a parabola is $f(x)=ax^2+bx+c$ , then

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

In the given function, a=-12 and b=120. So,

$-\dfrac{b}{2a}=-\dfrac{120}{2(-12)}$

$-\dfrac{b}{2a}=-\dfrac{120}{-24}$

$-\dfrac{b}{2a}=5$

Putting x=5 in the given function, we get

$P(5)=-12(5)^2+120(5)$

$P(5)=-12(25)+600$

$P(5)=-300+600$

$P(5)=300$

Therefore, 5 watt current will produce the maximum power of 300 amperes.

6 0

2 years ago