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3 years ago

A triangle has vertices (3,5), (5,2), and (2,0). Is the triangle a right triangle?

Mathematics

2 answers:

Artyom0805 [142]3 years ago

8 0

Answer:

No, this is not a right triangle

Step-by-step explanation:

katrin [286]3 years ago

7 0

No it is not a right triangle

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B) The perimeter of the shape is 25.71 cm

natita [175]

Answer:

$r=4cm$

Step-by-step explanation:

Having the radius to be involved, the shape is a circle.

<u>Circumference formula </u>

$c=2\pi r$

Finding radius, formula will be reversed.

$r=\frac{25.72cm}{2\pi }$

$r=\frac{25.71cm}{2(3.142)}$ $r=\frac{25.71cm}{6.284}$ $r=4.09cm$

4 0

3 years ago

Helpppppppppp me plz

creativ13 [48]

X+6 > -18+9x
18+6 > 8x
24 > 8x
3 > x

5 0

2 years ago

Seth buys last year's best-selling novel, in hardcover, for $17.85 $⁢17.85 . This is with a 15% 15% discount from the original p

Troyanec [42]

$21.00

let the original price = 100%

then the discount price = 100 - 15 = 85%

divide $17.85 by 85 to obtain 1% then multiply by 100 to obtain original price.

original price = $17.85 × $\frac{100}{85}$ = $21

8 0

2 years ago

Read 2 more answers

Find the value of x please help

dezoksy [38]

6x + 1 + 29 = 90

6x + 30 = 90

6x = 90 - 30

6x = 60

x = 60/6

x = 10

7 0

2 years ago

Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0

Alik [6]

Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

$(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0$

The general equation for the second-order polynomial is:

$x = \frac{2\cdot k \pm \sqrt{4\cdot k^{2}-4\cdot (k^{2}-1)\cdot (-3\cdot k + 1)}}{k^{2}-1}$

This condition must be observed for the case of a quadratic equation with equal roots:

$4\cdot k^{2} - 4\cdot (k^{2}-1)\cdot (-3\cdot k + 1) = 0$

$k^{2} + (k^{2}-1)\cdot (3\cdot k + 1) = 0$

$k^{2} + 3\cdot k^{3} - 3\cdot k - k^{2}-1 = 0$

$3\cdot k^{3} - 3\cdot k - 1 = 0$

7 0

3 years ago