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Nadusha1986 [10]

3 years ago

6

A shipping container will be used to transport several 70-kilogram crates across the country by rail. The greatest weight that c

an be loaded into the container is 23500 kilograms. Other shipments weighing 9700 kilograms have already been loaded into the container. What is the greatest number of 70-kilogram crates that can be loaded into the shipping container?

1 answer:

alina1380 [7]3 years ago

6 0

Subtract what has already been loaded from total amount allowed:

23500 - 9700 = 13,800

Divide the amount left by the weight of each crate:

13,800 / 70 = 197.14

They can load 197 crates.

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Directions: Calculate the area of a circle using 3.14x the radius

Leokris [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

8 0

2 years ago

The school librarian, Mr. Marker, knows the library has 1,400 books but wants to reorganize how the books are displayed on the s

kipiarov [429]

Answer:

200 resource books

800 fiction books

400 nonfiction books

Step-by-step explanation:

Let's define x as the amount of resource books that are in the library. As he has four times as many fiction books as resource books, that means he has 4*x fiction books.

We also know he has half as many nonfiction books as fiction books, that is 1/2*4*x = 2*x nonfiction books.

We just need to find out the value of x.

We know there are 1,400 books in total and there are just 3 types of books. Therefore, if we sum all of them, we get the 1,400.

Written in math language, that is

x + 4x + 2x = 1,400

7x = 1,400

We divide by 7 on both sides and we get

7/7x = 1,400/7

x = 200

Now we replace the x with 200 to know how many books he has of each type:

Resource books = x = 200

Fiction books = 4*x = 4*200 = 800

Nonfiction books = 2*x = 2*200 = 400

Therefore, there are 200 resource books, 800 fiction books and 400 nonfiction books in the library.

The tape diagram is attached as an image.

8 0

3 years ago

2 2/3 * 3/4 answer that

Nataly_w [17]

answer = 2

You can just use a calculator unless you need to show your work

5 0

3 years ago

The circumference of a circle is 65?. In terms of pi, what is the area of the circle?

iVinArrow [24]

Answer:

1056.25π square units

Step-by-step explanation:

A few formulas an definitions which will help us:

(1) $\pi=\frac{c}{d}$ , where c is the circumference of a circle and d is its diameter

(2) $A=\pi r^2$ , where A is the area of a circle with radius r. To put it in terms of d, remember that a circle's diameter is simply twice its radius, or mathematically, (3) $d=2r \rightarrow r=\frac{d}{2}$ .

We can rearrange equation (1) to put d in terms of π and c, giving us (4) $d = \frac{c}{\pi}$ , and we can make a few substitutions in (2) using (3) and (4) to get use the area in terms of the circumference and π:

$A=\pi r^2\\=\pi\left(\frac{d}{2}\right)^2\\=\pi\left(\frac{d^2}{4}\right)\\=\pi\left(\frac{(c/\pi)^2}{4}\right)\\=\pi\left(\frac{c^2/\pi^2}{4}\right)\\=\pi\left(\frac{c^2}{4\pi^2}\right)\\\\=\frac{\pi c^2}{4\pi^2}\\ =\frac{c^2}{4\pi}$

We can now substitute c for our circumference, 65, to get our answer in terms of π:

$A=\dfrac{65^2}{4\pi}=\dfrac{4225}{4\pi}=1056.25\pi$

8 0

3 years ago

Read 2 more answers

the stove you have to bake in is an old one and only has the temperature in ° Fahrenheit. You are making rusks and have to dry t

nalin [4]

Answer:

$80^{\circ} C$

Step-by-step explanation:

You are making rusks and have to dry them overnight at a temperature of 176° F.

The given temperature is in Fahrenheit. We need to convert it into °C.

The relation between Fahrenheit and °C is given by :

$C=\dfrac{5}{9}(F-32)$

Put F = 176

So,

$C=\dfrac{5}{9}(176-32)\\\\C=\dfrac{5}{9}\times 144\\\\C=80^{\circ} C$

So, the temperature is equal to $80^{\circ} C$ .

3 0

3 years ago