Please help me on this :-:

The legs of a right triangle measure 8 cm and 6 cm respectively. How long is the longest side of the triangle?

Masja [62]

Answer:

b. 10

Step-by-step explanation:

Based on the Pythagoras Theorem, the longest side of a right-angled triangle is: $\sqrt{a^{2}+b^{2} }$ where a and b are the legs of the triangle.

Longest side of triangle = $\sqrt{8^{2}+6^{2}}$

= $\sqrt{64+36}$

= $\sqrt{100}$

= 10

5 0

3 years ago

For a constant function, the y-coordinate is __________ for every pair of points selected, so the average rate of change always

sergij07 [2.7K]

In mathematics, a constant function is one defined in which a function whose output value (y) is the same for every value of the input variable (x). Therefore this means that whatever is the value for x, the value of y would always be the same. Hence we can say that the function is a straight horizontal line in a Cartesian plane.

Since the value of y does not change, and the average rate of change has the formula:

average rate of change = (y2 – y1) / (x2 – x1)

So y2 – y1 = 0 therefore average rate of change is also zero.

Hence to complete the sentence:

For a constant function, the y-coordinate is the same for every pair of points selected, so the average rate of change always has a value equal to zero.

4 0

3 years ago

Convert the angle 0 = 70 to radians. Express your answer exactly racians

sergeinik [125]

Answer:

$\frac{7\pi }{18}$

Step-by-step explanation:

$70*\frac{\pi }{180}$

$\frac{70\pi }{180}$

Simplify to the answer.

Hope this helped!

7 0

4 years ago

Will name you brainliest

Aleks [24]

Answer:

C po thanks

Step-by-step explanation:

sana po makatulong po sa inyo to

7 0

3 years ago

|x-3|<2 |4x+1|>0 |x-1|<5 Ayudaaaaaaa pleaseeeee

Dafna11 [192]

Recuerda que

• |x| = x si x ≥ 0

• |x| = -x si x < 0

Necesitas considerar dos casos:

• si x - 3 ≥ 0,

|x - 3| < 1 ⇒ x - 3 < 1 ⇒ x < 4

• si x - 3 < 0,

|x - 3| < 1 ⇒ -(x - 3) = 3 - x < 1 ⇒ -x < -2 ⇒ x > 2

Entonces la solución consta de todos los números reales x tales que x > 2 y x < 4, o simplemente 2 < x < 4.

El método para resolver las otras desigualdades es el mismo.

|4x + 1| > 0 ⇒ 4x + 1 > 0 o -(4x + 1) > 0

… ⇒ 4x + 1 > 0 o -4x - 1 > 0

… ⇒ 4x > -1 o -4x > 1

… ⇒ x > -1/4 o x < -1/4

⇒ x ≠ -1/4

|x - 1| < 5 ⇒ x - 1 < 5 o -(x - 1) < 5

… ⇒ x - 1 < 5 o -x + 1 < 5

… ⇒ x < 6 o -x < 4

… ⇒ x < 6 o x > -4

⇒ -4 < x < 6

7 0

3 years ago