All categories

3 years ago

Calculate the breath of the cuboid if the volume is 18 cubic cm. The height is 2cm and the width is 6cm

1 answer:

8 0

We know that:
$V=w \times h \times b$
Therefor:
$b=\frac{V}{h \times w}$
$b=\frac{16cm^{3}}{2cm \times 6cm}$
$b=\frac{18cm^{3}}{12cm^{2}}$
$b=1\frac{1}{2}cm=1.5cm$

You might be interested in

30 x 10 by the power of 2

Charra [1.4K]

Answer:

90,000

Step-by-step explanation:

30*10=300

300 to the second power is the same as 300*300

300*300=90,000

7 0

3 years ago

Read 2 more answers

Suppose a compact car uses 1 gallon of fuel for every 27 miles traveled. How would the graph for the compact car compare to the

laila [671]

You get 20 miles for every one gallon

8 0

3 years ago

Read 2 more answers

Based on her recipe, Elena knows that 5 servings of granola have 1,750 calories.

klemol [59]

Answer:

1,750 ÷5= 350

Step-by-step explanation:

there are 350 calories in each serving 1,750 is the total

4 0

3 years ago

Read 2 more answers

Model and partial quotient solve:<br><br> 783/4

vlada-n [284]

Answer:

195.75

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co

alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

$P=2x+2y$

On the other hand, since the available money to buy fence is D dollars,

$\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}$

Furthermore, the area of the enclosed land is given by

$A=xy$

Solving the second equation for x,

$\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}$

Substituting into the equation for the area,

$\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}$

To find the maximum possible area, solve A'(y)=0, as shown below

$\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}$

Therefore, the corresponding value of x is

$\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}$ <h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2> $\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}$ <h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

7 0

1 year ago