All categories

3 years ago

PUZZLE: six wolves catch six lambs in six minutes. How many wolves will be needed to catch sixty lambs in sixty minutes?

Mathematics

1 answer:

Korolek [52]3 years ago

4 0

Six i think cause it take each wolf 6 minutes to catch a lamb so it should still be six

You might be interested in

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers

A biscuit recipe calls for 2 pounds of all purpose flour and .75 pound of butter, what would be the baker’s percentage for butte

Goryan [66]

Answer: 37.5%

Explanation:

Percentage is defined as parts of an amount in 100 parts of a basis. This must be stated mathamatically for better understanding and for calculations. It is the ratio:

% = (parts / base) × 100.

In this case you need to express the amount of butter as a part in 100 parts of all purpose flour, which, using the above definition, is:

% butter = (0.75 pounds of butter / 2 pound of flour) × 100

% butter = 0.75 × 100 / 2 = 75 / 2 = 37.5

The answer is expressed as 37.5%.

4 0

3 years ago

Is the percent increase from 50 to 70 equal to the percent decrease from 70 to 50? Explain.

Galina-37 [17]

Answer

No. The amounts of change are the same, but the original amounts are different. The ratio for the percent increase from 50 to 70 is 20/50, or 40%. The ratio for the percent decrease from 70 to 50 is 20/70, or about 29%.

8 0

3 years ago

Read 2 more answers

Simplify the complex fraction.

lilavasa [31]

Given:

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}$

To find:

The simplified fraction.

Solution:

Step 1: Simplify the numerator

$$\frac{(4 r)^{3}}{15 t^{4}}=\frac{4^3 r^{3}}{15 t^{4}}=\frac{64 r^{3}}{15 t^{4}}$

Step 2: Simplify the denominator

$$\frac{16 r}{(3 t)^{2}}=\frac{16 r}{3^2 t^{2}}= \frac{16 r}{9 t^{2}}$

Step 3: Using step 1 and step 2

$$\frac{\left(\frac{(4 r)^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{(3 t)^{2}}\right)}=\frac{\left(\frac{64 r^{3}}{15 t^{4}}\right)}{\left(\frac{16 r}{9 t^{2}} \right)}$