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42  ÷  2

21  ÷  3

7    ÷  7

1

One way as follows:

7, 7, 1 gets an area of about 3.491.

Don't know whether thats the best though.

Typo, 5, 12, 13.

..... which is achieved when the triangle is a right triangle.  Why ?

Top line, a + b + c = 15,  so how can a = 5, b = 13, c = 13 be a solution ?

If R is the tension in the red rope and B is the tension in the blue rope, then resolving horizontally,

\(R\cos\theta = B\)

and resolving vertically

\(R\sin\theta = 1.25g.\)

Dividing the second equation by the first,

\(\tan\theta=1.25g/B, \text{ so } B=1.25g/\tan\theta.\)

Let the sides of the triangle be $a$, $b$, and $c$. Since the perimeter is 15cm, we have $a+b+c=15$. We want to find the smallest possible area of the triangle, which is achieved when the triangle is a right triangle. Therefore, we will try to find a Pythagorean triple $(a,b,c)$ that satisfies $a+b+c=15$.

Since $a^2+b^2=c^2$, we have $a+b+c=\sqrt{a^2+b^2}+c$, which implies $c = 15 - \sqrt{a^2+b^2}$. Substituting this into the Pythagorean equation, we get

$$a^2+b^2 = (15-\sqrt{a^2+b^2})^2 = 225-30\sqrt{a^2+b^2}+a^2+b^2.$$

Simplifying, we get

$$30\sqrt{a^2+b^2}=225 \qquad\Rightarrow\qquad \sqrt{a^2+b^2}=\frac{225}{30}=\frac{15}{2}.$$

Since $a$ and $b$ are integers, we want to find Pythagorean triples with hypotenuse $\frac{15}{2}$. The only such triple is $(5,12,13)$. Therefore, the sides of the triangle are $a=5$, $b=12$, and $c=8$, and the area is

$$\frac{1}{2}ab = \frac{1}{2}(5)(12) = 30.$$

Therefore, the smallest possible area for a triangle of integral side lengths whose perimeter is 15cm is $\boxed{30}$ square centimeters.

If Jackie's taxable income for 2020 is less than $9,875, the equation to represent the amount of income tax that Jackie owes, T, in terms of her taxable income, J, would be T = 0.10J. This is because the tax rate for taxable income up to $9,875 for single filers in 2020 was 10%.

By the quadratic formula, the solutions are \(\dfrac{7 \pm i \sqrt{7}}{4}\)

Then \(a^3 + b^3 = \left( \dfrac{7 + i \sqrt{7}}{4} \right)^3 + \left( \dfrac{7 - i \sqrt{7}}{4} \right)^3 = \dfrac{49}{8}\)

The length BC is 51/5.

Let's start with what we have:

-4x + 6(x + 1)

This simplifies to:

-4x + 6x + 6

Which is equal to 

 2x + 6 

This is the volume of a Frustrum:

Volume of frustum of a rectangular pyramid = h/3 * (l*w + sqrt(l*w*l1*w1) + l1*w1)

Length l: 
202

Breadth w: 
25

Length l1: 
180

Breadth w1: 
20

Height h: 
15

The calculated volume is: 64,569.0056 cu feet.

Area of the top =202  x  25 = 5050 ft^2

Area of the bottom=180  x  20 = 3600 ft^2

Average area of the bottom and the top:

(5050  +  3,600) / 2  = 4,325 ft^2

Volume of earth removed:

15  x  4,325 =64,875 ft^3 - volume of earth to be removed.

(1, 2) ,(1, 4) ,(1, 6) ,(2, 1) ,(2, 2) ,(2, 3) ,(2, 4) ,(2, 5) ,(2, 6) ,(3, 2) ,(3, 4) ,(3, 6) ,(4, 1) ,(4, 2) ,(4, 3) ,(4, 4) ,(4, 5) ,(4, 6) ,(5, 2) ,(5, 4) ,(5, 6) ,(6, 1) ,(6, 2) ,(6, 3) ,(6, 4) ,(6, 5) ,(6, 6) ,>>Total =  27 ways.

The probability is: 27 / 6^2 ==27 / 36 ==3 / 4

3960  x 2 ==7,920 miles - Earth's diameter at the equator.

Circumference ==Diameter  x  Pi

Circumference ==7,920  x  3.141592 ==24,881.41 - miles at the equator

Speed ==Distance (Circumference) / Time

4/3600 + 56 / 60 + 23 ==23.93444444......hours taken for 1 full rotation of Earth at equator.

Speed ==24,881.41 / 23.93444..

Speed ==1,039.56 miles per hour.

8x^2 - 6x=0

Factor the left-hand side

2x [4x  -  3] ==0

2x ==0

x ==0

[4x  -  3]==0

4x == 3

x ==3 / 4

Hi Guest,

\(2x^2=\sqrt{2}x \implies x(2x-\sqrt{2})=0\)

So: \(x_1=0 \\ x_2=\dfrac{\sqrt{2}}{2} \)

I hope this helps.

\(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3},\)

so subtracting the second equation from the first,

\(\displaystyle (a-b)^{3}=679 -615=64,\)

so

\(\displaystyle a-b=4.\)