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3 years ago

Is a triangle with side lengths of 4,5, and 6 inches a right triangle?

Mathematics

1 answer:

oee [108]3 years ago

5 0

Answer:

yes it can make a right triangle

Step-by-step explanation:

You might be interested in

What are whole number dimensions?

Sphinxa [80]

When you get to the point where the width is larger than the length, you can stop because you have exhausted all the possibilities. The greatest area will be the rectangle that is closer in to a square than any of the others.

5 0

3 years ago

Read 2 more answers

I Need Help Like Now!!

kogti [31]

Answer:

$\frac{136}{999}$

Step-by-step explanation:

We require 2 equations with the repeating decimal after the decimal point in both equations.

The bar above 136 indicates that 136 is being repeated

let x = 0.136136...... → (1) Multiply both sides by 1000

1000x = 136.136.... → (2)

Subtract (1) from (2) to eliminate the decimal part

999x = 136 ( divide both sides by 999 )

x = $\frac{136}{999}$

5 0

2 years ago

What is the measure of an angle if it is 260 degrees less than six times its own complement?

poizon [28]

Answer:

40 degrees

Step-by-step explanation:

Complementary angles have a sum of 90.

y = 90 - x

The angle is 260 degrees less than six times its complement.

y = 6x - 260

90 - x = 6x - 260

350 - x = 6x

350 = 7x

50 = x

This is the measure of the complement.

y = 90 - x

y = 90 - 50

y = 40

This is the measure of the first angle.

7 0

3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

2 years ago

What is the common denominator for 32 and 3

Leya [2.2K]

Just multiply 32 by 3

4 0

3 years ago