All categories

1 year ago

3. Ivan has 15 yards of green felt and 12 yards of

Mathematics

2 answers:

lys-0071 [83]1 year ago

6 0

Answer: 5 yards of green felt and 4 yards of blue felt is used to make each quilt.

Step-by-step explanation:

As per given ,

Length of green felt used to make 3 quilts = 15 yards

Then , the length of green felt used to make 1 quilt = $\dfrac{15}{3}=5\text{ yards}$

Length of blue felt used to make 3 quilts = 12 yards

Then , the length of blue felt used to make 1 quilt = $\dfrac{12}{3}=4\text{ yards}$

Hence, 5 yards of green felt and 4 yards of blue felt is used to make each quilt.

dybincka [34]1 year ago

4 0

Answer:

three yards for blue felt and five yards for green

Step-by-step explanation:

divide each by three

You might be interested in

se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y

alexandr1967 [171]

Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0

2 years ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago

Gigi earned $65 for 5 hours of gardening. She earned $90 for 9 hours of office work. Which statement correctly compares Gigi’s e

hjlf

She earned $3 more per hour for gardening than for office work.

8 0

2 years ago

Read 2 more answers

The vet told Jake that his dog, Rocco, who weighed 55 pounds, needed to lose 10 pounds. Jake started walking Rocco every day and

madreJ [45]

Answer:it would take him five weeks to lose ten pounds

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Clint and Max worked out at the gym together today. Clint works out at the gym every 4 days. Max works out at the gym every 6 da

VikaD [51]

Answer:

24 days

Step-by-step explanation:

To determine the number of days until they both work out at the gym on the same day again,

Find the lowest common multiples of 4 days and 6 days

Clint (4 days) = 8, 12, 16, 20, 24, 28, 32

Max (6 days) = 12, 18, 24, 30, 36, 42

The lowest common multiple of ,4 days and 6 days is 24 days

Therefore, the number of days until they both work out at the gym on the same day again is 24 days

7 0

1 year ago