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3 years ago

Ken can walk 40 dogs in 8 hours. How many dogs can Ken walk in 12 hours?

Mathematics

1 answer:

kompoz [17]3 years ago

7 0

40 dogs in 8 hours
divide by 8
5 dogs in 1 hour (12 minutes per dog, wow!)
12 hours
times 12
60 dogs in 12 hours

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Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering p

tankabanditka [31]

Answer:

1.778 times more or 16/9 times more

Step-by-step explanation:

Given:

- Mirror 1: D_1 = 8''

- Mirror 2: D_2 = 6"

Find:

Compare the light gathering power of an 8" primary mirror with a 6" primary mirror. The 8" mirror has how much light gathering power?

Solution:

- The light gathering power of a mirror (LGP) is proportional to the Area of the objects:

LGP ∝ A

- Whereas, Area is proportional to the squared of the diameter i.e an area of a circle:

A ∝ D^2

- Hence, LGP ∝ D^2

- Now compare the two diameters given:

LGP_1 ∝ (D_1)^2

LGP ∝ (D_2)^2

- Take a ratio of both:

LGP_1/LGP_2 ∝ (D_1)^2 / (D_2)^2

- Plug in the values:

LGP_1/LGP_2 ∝ (8)^2 / (6)^2

- Compute: LGP_1/LGP_2 ∝ 16/9 ≅ 1.778 times more

6 0

3 years ago

Will mark Brainliest!! :)

Georgia [21]

Step-by-step explanation:

The system of equations for eq 1 which is 3x + y = 118 represents the Green High School which filled three buses(with a specific number of students identified as x) and a van(with a specific number of students identified as y) with a total of 118 students.

for eq 2; 4x + 2y = 164; represents Belle High School which filled four buses(with a specific number of students identified as x) and two vans(with a specific number of students identified as y) with a total of 164 students.

The solution represents the specific number of students in the buses and vans in eq1 and eq 2 with x being 36 students and y being 10 students.

substituting 36 for x and 10 for y in eq 1;

3(36) + 10 = 108 + 10 = 118 total students for Green High School

substituting 36 for x and 10 for y in eq2;

4(36) + 2(10) = 144 + 20 = 164 total students for Belle High school

6 0

2 years ago

7 to the negative 3rd porwer times 7 to the negative 2nd porwer times 2 the 0 power divided by 7 to negative 1st power times 7 t

Jobisdone [24]

oof

remember some rules

(a/b)(c/d)=(ac)/(bd)

$\frac{x^a}{x^b}=x^{a-b}$

$x^0=1$

$(x^a)(x^b)=x^{a+b}$

$x^{-a}=\frac{1}{x^a}$

$\frac{(7^{-3})(7^{-2})(2^0)}{(7^{-1})(7^{-4})(2^3)}=$

$\frac{(7^{-3-2})(2^0)}{(7^{-1-4})(2^3)}=$

$\frac{(7^{-5})(2^0)}{(7^{-5})(2^3)}=$

$(\frac{7^{-5}}{7^{-5}})(\frac{2^0}{2^3})=$

$(7^{-5-(-5)})(2^{0-3})=$

$(7^0)(2^{-3})=$

$\frac{1}{2^3}=$

$\frac{1}{8}$

7 0

3 years ago

A bakery sold 16 apple pie on Saturday they sold 4 more apple pies on Sunday than they did on Saturday how many apple pies did t

lys-0071 [83]

Answer:

20 apple pies

Step-by-step explanation:

On Saturday they sold 16, while on Sunday they sold 4 more.

16+4 = 20.

20 apple pies.

4 0

3 years ago

The equation D=13/v shows that the density of a particular substance equals a mass of 13 grams divided by the volume of the subs

lions [1.4K]

For this case we have the following equation:
$D = \frac{13}{v}$
Where,
D: the density of a particular substance
v: the volume of the substance
When replacing v = 0 in the given equation we have:
$D = \frac{13}{0}$
$D = \infty$
This means that as the function acquires values very close to zero, the density acquires a very large value.
Answer:
as the volume approaches 0:
 The density approaches infinity.
 option 1

8 0

3 years ago

Read 2 more answers