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2 years ago

Identify the sum of the interior angle measures of a convex octagon. HELP PLEASE ASAP!!

Mathematics

1 answer:

skad [1K]2 years ago

7 0

Answer:

Because the sum of these angles will always be 360°, then each exterior angle would be 60° (360° ÷ 6 = 60°). If each exterior angle is 60°, then each interior angle is 120° (180° − 60° = 120°).

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Choose the inequality that could be used to solve the following problem.

yulyashka [42]

Answer:

3x ≥ -6

Step-by-step explanation:

"No less than" means "greater than or equal to". An appropriate translation of the problem statement is ...

3x ≥ -6

8 0

2 years ago

Read 2 more answers

How is the graph of y=^3√0.5x related to its parent function,y=^3√x ? A.It is horizontally stretched by a factor of 0.5 B. It is

uranmaximum [27]

For this case we have the following function:
y = ^ 3√x
Applying the following transformation we have:
Expansions and horizontal compressions
The graph of y = f (bx):
If 0 <b <1, the graph of y = f (x) expands horizontally by the factor of 1 / b.
y = ^ 3√0.5x
Answer:
A.It is horizontally stretched by a factor of 1 / 0.5

8 0

3 years ago

How do you solve x+2a=16+ax-6a ?

nasty-shy [4]

X+2a = 16+aX-6a
8X - 16 = aX-2a
8(X-2) = a(X-2)
(X-2)/(X-2) = a/8
1 = a/8
a= 8

3 0

2 years ago

4 ( x- 2) = 34 for x

DerKrebs [107]

X=21/2
X=10 1/2
X=10.5

7 0

2 years ago

Read 2 more answers

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago