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2 years ago

(I will give brainliest. Please answer quickly)

Mathematics

1 answer:

ladessa [460]2 years ago

4 0

The first step to solve this problem is to find the area of the rectangular piece of fabric.

A of triangle = bh/2

A = (14 cm) (6 cm) /2

A = 84 cm^2 / 2

A = 42 cm

And since there are 31 pieces of the fabric, the total area of all the pieces of fabric is:

31 pieces of fabric x 42 square centimeters per piece = 1,302 square centimeters

To computer how many congruent triangular patches can be cut, you have to divide the total area of the fabric pieces with the area of the congruent triangle:

1,302 square centimeters / 21 square centimeters = 62

Therefore, Leia can cut 62 patches.

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2 years ago

Sam purchases a new car for $29,500. The car depreciates at a rate of 13.25% per year. What is the value of the car after 7 year

miv72 [106K]

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3 years ago

Principal=2000,R=12percent,Time=2years and 6 months.

almond37 [142]

Principal=P=2000
Rate of interest=12%
Time=2years and 6months=2.5years
Interest=I

$\boxed{\sf I=\dfrac{PRT}{100}}$

$\\ \sf\longmapsto I=\dfrac{2000(12)(2.5)}{100}$

$\\ \sf\longmapsto I=\dfrac{24000(2.5)}{100}$

$\\ \sf\longmapsto I=\dfrac{60000}{100}$

$\\ \sf\longmapsto I=600$

7 0

1 year ago

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A piece of wire 8 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral trian

padilas [110]

Answer:

a. 8

b. 7

Step-by-step explanation:

Let $x$ be the length of wire for the square and $y = 8 -x$ (eq. 1) be the length of the wire for the equilateral triangle.

We know the $x$ is the total length of the wire used, hence the length of a side of the square will be $\frac{x}{4}$ . The area of the square will be: $(\frac{x}{4}) ^ 2$

Similarly for the equilateral triangle, perimeter is $y$ . Hence the area will be $\frac {\sqrt {3}\ y^2}{4}$ .

The total area of both shapes will be: $A = (\frac{x}{4})^2 + \frac{\sqrt {3} \ y^2}{4}$

We will substitute the value of y from eq. 1:

$A = (\frac{x}{4})^2 + \frac{\sqrt {3} \ (8-x)^2}{4}$

We find the derivative of the above function to find maximum and minimum: $f'(x) = 0$ ⇒ $A'(x) = 0$

$A' = \frac{x}{8} - \frac{\sqrt {3} \ (8-x)}{2}$

$A' = \frac{x}{8} - \frac{\sqrt {3} \ (8-x)}{2} = 0\\\\\frac{x}{8} - \frac{8\sqrt{3} -\sqrt{3} x}{2} =0\\\\x - 4(8\sqrt{3} -\sqrt{3}x)=0\\\\x-32\sqrt3 - 4\sqrt3 x = 0\\\\x - 4\sqrt3 = 32 \sqrt3 \\\\(1 - 4\sqrt3)x = 32 \sqrt3 \\\\x = \frac{32 \sqrt3 }{(1 - 4\sqrt3)}\\\\x =6.99 \approx 7.00$

We find $A''(7)$ to check whether $x = 7$ is the minimum or maximum of the function.

$A'' = \frac {1}{8} +\frac {\sqrt {3}}{2} \\\\A'' = 0.99 \approx 1.00$

Hence, $x = 7$ is the minimum and $x = 8$ will be the maximum

8 0

2 years ago