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1 year ago

Classify the triangle by its angles and sides

Mathematics

1 answer:

postnew [5]1 year ago

7 0

Answer; A: Right scalene. Square on the corner tells you it's a right triangle. It's a scalene triangle because none of the sides are equal.

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PLEASE ANSWER QUICK!!

Citrus2011 [14]

Answer:

A) B is greater

B) when C = 21, A = 5 and B = 15

C) A must be 0, 5, or 12 for C to be greater than B

Step-by-step explanation:

A) when a = -3, c = -20 and b = 4(-3)-5 so B = -17

B) when c = 21, a = 5 and b = 4(5)-5 so B = 15

C) C is greater than B when a = 0, 5, and 12

when a = 0, b = 4(0)-5 or -5

when a = 5, b = 4(5)-5 or 15

when a = 12, b = 4(12)-5 or 43

3 0

2 years ago

Which part of the water cycle is responsible for cloud formation

julia-pushkina [17]

It’s Condensation !!

6 0

1 year ago

This exercise involves the formula for the area of a circular sector. The area of a circle is 700 m2. Find the area of a sector

mezya [45]

Answer:

334.4 m²

Step-by-step explanation:

The formula for the area of a sector is given as:

1/2 × r² × θ

Where θ = Central angle

Area of a Circle = 700 m²

The formula for the area of a circle = πr²

r = Radius of a circle

r² = Area / π

r = √Area / π

r = √700/π

r = 14.927053304 m

Approximately, r = 14.93 m

Therefore, the area of the sector

= 1/2 × r² × θ

= 1/2 × 14.93² × 3 rad

= 334.35735 m²

Approximately, Area of the sector = 334.4 m²

4 0

2 years ago

Madison worked her 40 regular hours last week plus 12 overtime hours at the time and a half rate. Her gross pay was $842.00. Wha

GrogVix [38]

she has $13.80 every hour because 842 divided by 40 plus 12 equals 61 so then and you cant round anymore.

3 0

2 years ago

In a triangle ABC, measure of angle B is 90 degrees. AB is 3x-2 units and BC is x+3. If the area of the triangle is 17 sq cm, fo

Scorpion4ik [409]

Answer:

$x=\frac{8}{3}\ cm$

Step-by-step explanation:

we know that

The area of the right triangle ABC is equal to

$A=\frac{1}{2}(AB)(BC)$

we have

$A=17\ cm^2$

$AB=(3x-2)\ cm$

$BC=(x+3)\ cm$

substitute the values

$17=\frac{1}{2}(3x-2)(x+3)$

$34=(3x-2)(x+3)$

$34=3x^2+9x-2x-6$

$3x^2+7x-6-34=0$

$3x^2+7x-40=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^2+7x-40=0$

$a=3\\b=7\\c=-40$

substitute in the formula

$x=\frac{-7(+/-)\sqrt{7^{2}-4(3)(-40)}} {2(3)}$

$x=\frac{-7(+/-)\sqrt{529}} {6}$

$x=\frac{-7(+/-)23} {6}$

$x=\frac{-7(+)23} {6}=\frac{16}{6}=\frac{8}{3}$

$x=\frac{-7(-)23} {6}=-5$

therefore

The solution is

$x=\frac{8}{3}\ cm$

6 0

2 years ago