All categories

2 years ago

Ella uses 3 inches of ribbon for each key chain she makes. How many key chains can she make with 4 yards of ribbon?

Mathematics

1 answer:

Marina CMI [18]2 years ago

8 0

Answer:

Step-by-step explanation:

You might be interested in

Algebra- pls help ;-;<br> will mark brainliest

Pavlova-9 [17]

Answer:

The solution is correct.

7 0

2 years ago

Read 2 more answers

Help pls need answer

blagie [28]

Answer:

Step-by-step explanation:

idk

4 0

2 years ago

A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of t

Makovka662 [10]

Answer:

132.233 ft2

Step-by-step explanation:

Let's call the width of the rectangle 'w' and the length 'x'. So the area of the semicircle is:

$A_1 = \pi*radius^2/2$

$A_1 = \pi*(w/2)^2/2$

$A_1 = \pi/8*w^2$

And the area of the rectangle is:

$A_2 = w*x$

If the perimeter of the window is 41 feet, we have:

$Perimeter = length + 2*width + \pi*radius$

$41 = x + 2*w + \pi*w/2$

$x = 41 - w(2 + \pi/2)$

Now, the equation for the total area of the window is:

$A = A_1 + A_2 = \pi/8*w^2 + w*x$

$A = \pi/8*w^2 + w*(41 - w(2 + \pi/2))$

$A = (\pi/8-2 - \pi/2)*w^2 + 41w = -3.1781w^2 + 41w$

To find the maximum area, we can find the x-coordinate of the vertex of the quadratic equation:

$x\_vertex = -b / 2a = -41 / (-3.1781*2) = 6.45$

So the width that gives us the maximum area of the window is 6.45 feet, and the area will be:

$A = -3.1781w^2 + 41w = -3.1781*(6.45)^2 + 41*6.45 = 132.233\ ft^2$

3 0

2 years ago

If a cup of coffee has temperature 97Â°C in a room where the ambient air temperature is 25Â°C, then, according to Newton's Law o

FinnZ [79.3K]

Answer:

The average temperature is $T_{a} = 81.95^oC$

Step-by-step explanation:

From the question we are told that

The temperature of the coffee after time t is $T(t) = 25 + 72 e^{[-\frac{t}{45} ]}$

Now the average temperature during the first 22 minutes i.e fro $0 \to 22$ minutes is mathematically evaluated as

$T_{a} = \frac{1}{22-0} \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx$

$T_{a} = \frac{1}{22} [25 t + 72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.$

$T_{a} = \frac{1}{22} [25 t - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.$

$T_{a} = \frac{1}{22} [25 (22) - 3240e^{[-\frac{22}{45} ]} - (- 3240e^{0} )]$

$T_{a} = \frac{1}{22} [550 - 1987.12 + 3240]$

$T_{a} = 81.95^oC$

8 0

2 years ago

PLEASE HELP ASAP

balandron [24]

Answer:

Step-by-step explanation:

X is the variable and x means multiplication symbol.

X + (X x 3 - 1) + (X x 3 - 1) - 4) ≥ 45 is the answer I got.

The answer to that inequality would be

X ≥ 7 I think

7 0

2 years ago