All categories

3 years ago

What is the largest possible area of a right triangle whose hypotenuse is 20?

Mathematics

2 answers:

Monica [59]3 years ago

5 0

Answer:the largest possible area for a right triangle whose hypotenuse is 5 cm long is area =6.25cm^2

Step-by-step explanation:

kherson [118]3 years ago

3 0

Answer:

171

Step-by-step explanation:

Since the hypotenuse is the longest side on a right triangle, the other two sides would have to be less than 20. So, you would get the two numbers under 20, which are 19 and 18. In order to find the area of a right triangle you do A x B x 1/2. So you would do 18 x 19 x 1/2 which is 171.

You might be interested in

Please help:) i would appreciate it lol

Svetllana [295]

Answer:

subtotal: $65.00

8% sales tax: $5.20

total: $70.20

Step-by-step explanation:

6 0

3 years ago

Solve for n 20 > -4x - 8

LekaFEV [45]

The answer is X>-7 for your question

5 0

3 years ago

In the expansion (ax+by)^7, the coefficients of the first two terms are 128 and -224, respectively. Find the values of a and b

madam [21]

Answer:

a = 2, b = 3.5

Step-by-step explanation:

Expanding $(ax+by)^7$ using Binomial expansion, we have that:

$(ax+by)^7$ =

$(ax)^7(by)^0 + (ax)^6(by)^1 + (ax)^5(by)^2 + (ax)^4(by)^3 + (ax)^3(by)^4 + (ax)^2(by)^5 + (ax)^1(by)^6 + (ax)^0(by)^7$

$= (a)^7(x)^7+ (a)^6(x)^6(b)(y) + (a)^5(x)^5(b)^2(y)^2 + (a)^4(x)^4(b)^3(y)^3 + (a)^3(x)^3(b)^4(y)^4 + (a)^2(x)^2(b)^5(y)^5 + (a)(x)(b)^6(y)^6 + (b)^7(y)^7\\\\\\= (a)^7(x)^7+ (a)^6(b)(x)^6(y) + (a)^5(b)^2(x)^5(y)^2 + (a)^4(b)^3(x)^4(y)^3 + (a)^3(b)^4(x)^3(y)^4 + (a)^2(b)^5(x)^2(y)^5 + (a)(b)^6(x)(y)^6 + (b)^7(y)^7$

We have that the coefficients of the first two terms are 128 and -224.

For the first term:

=> $a^7 = 128$

=> $a = \sqrt[7]{128}\\ \\\\a = 2$

For the second term:

$a^6b = -224$

$b = \frac{-224}{a^6}$

$b = \frac{-224}{2^6} \\\\\\b = \frac{-224}{64} \\\\\\b = 3.5$

Therefore, a = 2, b = 3.5

5 0

3 years ago

Help me with solving multi-step equations...13x + 2 = 4x + 38

svet-max [94.6K]

1 Subtract 22 from both sides
13x=4x+38-213x=4x+38−2

2 Simplify 4x+38-24x+38−2 to 4x+364x+36
13x=4x+3613x=4x+36

3 Subtract 4x4x from both sides
13x-4x=3613x−4x=36

4 Simplify 13x-4x13x−4x to 9x9x
9x=369x=36

5 Divide both sides by 99
x=\frac{36}{9}x=
9

36

6 Simplify \frac{36}{9}
9

36
to 44
x=4x=4

3 0

3 years ago

Read 2 more answers

Which statement about the two triangles is correct

Bess [88]

Answer:

They might be the same type of triangle

Step-by-step explanation:

8 0

3 years ago