Divide.<br><br> 1.61÷0.35<br><br> Enter your answer in the box.

Jonathan is designing a rectangular path for his garden using paving stones. The paving stones are square and 50cm wide. If the

koban [17]

Answer: 32 paving stones

Step-by-step explanation:

Hi, to answer this question, first we have to convert the measures of the paving stones into meters.

Since: 1m = 100 cm

50 cm /100 = 0.5m

Now, we have to calculate the area of each paving stone:

Area of a square: s^2 = 0.5^2 = 0.25m2

Next, the area of the path.

Area of a rectangle = width x length = 1 x 8 = 8 m2

Finally, we have to divide the area of the path by the area of each paving stone:

8/0.25 = 32 paving stones

Feel free to ask for more if needed or if you did not understand something.

8 0

3 years ago

Read 2 more answers

Help, math (•_•) …........I have to put a lot of words but the question is on the picture

Marta_Voda [28]

What do u wanna know

8 0

3 years ago

Ar a furniture store, a bed frame costs $440 and mattress costs $654. There are currently 7 bed frames in the store, and togethe

Gnoma [55]

Answer:

the three angles of a triangle have measures x,2x,and 4y,where x>56.If x and y are integers, what is one possible value of y?

6 0

2 years ago

Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y

professor190 [17]

Answer:

Required total charge is $256\pi$ coulombs per square meter.

Step-by-step explanation:

Given electric charge is dristributed over the disk,

$x^2=y^2\leq 16$ so that the charge density at (x,y) is,

$\rho (x,y)=2x+2y+2x^2+2y^2$

To find total charge on the disk let Q be the total charge and $x=r\cos\theta,y=r\sin\theta$ so that,

$Q={\int\int}_Q\rho(x,y) dA$ where A is the surface of disk.

$=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA$

$=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta$

$=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta$

$=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta$

$=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta$

$=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi$

$=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi$

$=256\pi$

Hence total charge is $256\pi$ coulombs per square meter.

3 0

3 years ago

the slope of a line passing through h (-2, 5) is -3/4. Which ordered pair represents a point on this line?

rosijanka [135]

Answer:

A

Step-by-step explanation:

Calculate the slope of the given points with the point (- 2, 5 )

If the slope is - $\frac{3}{4}$ then the point is on the line

Calculate slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (6, - 1)

m = $\frac{-1-5}{6+2}$ = $\frac{-6}{8}$ = - $\frac{3}{4}$ ← point (6, - 1) is on the line

Repeat with (x₁, y₁ ) = (- 2, 5 ) and (x₂, y₂ ) = (2, 8)

m = $\frac{8-5}{2+2}$ = $\frac{3}{4}$ ← point (2, 8) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (- 5, 1)

m = $\frac{1-5}{-5+2}$ = $\frac{-4}{-3}$ = $\frac{4}{3}$ ← point (- 5, 1) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (1, 1)

m = $\frac{1-5}{1+2}$ = - $\frac{4}{3}$ ← point (1, 1) is not on the line

Thus the point on the line is (6, - 1 ) → A

4 0

3 years ago

Divide. 1.61÷0.35 Enter your answer in the box.