Ask question

Ask question

All categories

2 years ago

8

Four people at pia’s pottery shop each make 29 mugs and 18 pottery bowls three people at Jason’s craft shop each make the same n

umber of mugs and twice as many pottery bowls how many objects did seven people make in all

1 answer:

BartSMP [9]2 years ago

8 0

The seven people made 383 objects in all

Step-by-step explanation:

The given is:

Four people at pia’s pottery shop each make 29 mugs and 18 pottery bowls
Three people at Jason’s craft shop each make the same number of mugs and twice as many pottery bowls

We need to find how many objects the seven people made in all

∵ Four people at pia’s pottery shop each make 29 mugs

∵ Three people at Jason’s craft shop each make the same

number of mugs

∴ Seven people in the two shops made 29 mugs each

∴ The total number of mugs = 7 × 29

∴ The total number of mugs = 203

∵ Four people at pia’s pottery shop each make 18 pottery bowls

∵ Three people at Jason’s craft shop each make twice as

many pottery bowls

- Multiply the number of bowls by 2

∴ Three people at Jason’s craft shop each make 2 × 18 = 36

pottery bowls

∴ The total number of pottery bowls = (4 × 18) + (3 × 36)

∴ The total number of pottery bowls = 72 + 108

∴ The total number of pottery bowls = 180

Now add the total number of mugs and the total number of pottery bowls to find the total number of objects the seven people made

∵ The total number of mugs is 203

∵ The total number of pottery bowls is 180

∴ The total number of objects = 203 + 180

∴ The total number of objects = 383

The seven people made 383 objects in all

Learn more;

You can learn more about the word problems in brainly.com/question/10557938

#LearnwithBrainly

You might be interested in

On a map, Cary drew a set of coordinate axes. He noticed that the town of Coyner is located at the point (3, 1) and Woottonville

sp2606 [1]

Answer:

The coordinates of the school should be at (10.5 , 4.25)

Step-by-step explanation:

We are dealing with two points here on a plane. These are:

Town of Coyner, Point A(3,1)

town of Coyner, Point B (18, 7.5)

in coordinate geometry, to get the mid point of two points we simply use the formula:

$(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$

where x and y are the first and second points respectively

applying this formula, we have that

the mid point of the two towns is

$(\frac{3+18}{2}, \frac{1+7.5}{2})=(10.5 , 4.25)$

Hence the coordinates of the school should be at (10.5 , 4.25)

4 0

2 years ago

Find the volume of a right circular cone that has a height of 12.5 cm and a base with a circumference of 5.8 cm. Round your answ

BaLLatris [955]

Answer:

$V = 11.2 cm^3$

Step-by-step explanation:

Given

Shape: Right Circular cone

Height, h = 12.5 cm

Base circumference, C = 5.8cm

Required:

Calculate Volume

The volume of a cone is calculated as thus;

Volume, $V = \frac{1}{3}\pi r^2h$

Where V, r and h represent the volume, the radius and the height of the cone respectively

But first we need to calculate the radius of the cone.

Using formula of circumference.

$C = 2\pi r$

We have C to be 5.8cm

By substituting this value;

$5.8 = 2\pi r$

Divide through by $2\pi$

$\frac{5.8}{2\pi} = \frac{2\pi r}{2\pi}$

$r = \frac{5.8}{2\pi}$

$r = \frac{2.9}{\pi}$

Now, we can calculate the volume of the cone

By substituting $r = \frac{2.9}{\pi}$ and h = 12.5 cm;

$V = \frac{1}{3}\pi r^2h$ becomes

$V = \frac{1}{3}\pi * (\frac{2.9}{\pi})^2 * 12.5$

$V = \frac{1}{3}\pi * \frac{8.41}{\pi^2} * 12.5$

$V = \frac{1}{3} * \frac{8.41}{\pi} * 12.5$

Take $\pi$ as 3.14

$V = \frac{1}{3} * \frac{8.41}{3.14} * 12.5$

$V = 11.15976$

$V = 11.2 cm^3$ ---- Approximated

Hence, the volume of the cone is; $V = 11.2 cm^3$

4 0

2 years ago

A country's population in 1993 was 94 million. in 1999 in was 99 million. estimate the population in 2005 using the exponential

Jet001 [13]

The initial population is
P₀ = 94 million in 1993

The growth formula is
$P(t) = P_{0}e^{kt}$
where P(t) is the population (in millions) after t years, measured from 1993.
k = constant.

Because P(5) = 99 million (in 1999),
$94e^{5k} = 99 \\e^{5k}=1.0532 \\ 5k = ln(1.0532) \\ k = 0.010367$

In the year 2005, t = 12 years, and
$P(12)=94e^{0.010367*12} = 106.45$

Answer: 106 million (nearest million)

7 0

2 years ago

The measure of the supplement of an angle exceeds three times the measure of the complement by 30. Find the measure of the angle

USPshnik [31]

Answer:

Angle = 45

Supplement = 135

Complement = 45

Step-by-step explanation:

5 0

2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago