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4 years ago

Please help me ASAP!!!!

Mathematics

1 answer:

son4ous [18]4 years ago

3 0

$Answer: \\ 14. \: 5x = 180°\Leftrightarrow x = 36° \\ 15. \: 10x = 90°\Leftrightarrow x = 9°$

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Can someone answer this for me? I don’t understand how to do i.

LiRa [457]

These are the steps good luck

7 0

3 years ago

How do you know a radical expression is in simplest form?

Diano4ka-milaya [45]

Answer: To know whether a radical expression is in simplest form or not you should put the numbers and letters inside the radical in terms of prime factors. Then, the radical expression is in the simplest form if all the numbers and letters inside the radical are prime factors with a power less than the index of the radical

Explanation:

Any prime factor raised to a power greater than the index of the root can be simplified and any factor raised to a power less than the index of the root cannot be simplified

For example simplify the following radical in its simplest form:

$\sqrt[5]{3645 a^8b^7c^3}$

1) Factor 3645 in its prime factors: 3645 = 3^6 * 5

2) Since the powr of 3 is 6, and 6 can be divided by the index of the root, 5, you can simplify in this way:

- 6 ÷ 5 = 1 with reminder 1, so 3^1 leaves the radical and 3^1 stays in the radical

3) since the factor 5 has power 1 it can not leave the radical

4) the power of a is 8, then:

8 ÷ 5 = 1 with reminder 3 => a^1 leaves the radical and a^3 stays inside the radical.

5) the power of b is 7, then:

7 ÷ 5 = 1 with reminder 2 => b^1 leaves the radical and b^2 stays inside the radical

6) the power of c is 3. Since 3 is less than 5 (the index of the radical) c^3 stays inside the radical.

7) the expression simplified to its simplest form is

$3ab \sqrt[5]{3.5.a^3b^2c^3}$

And you know it cannot be further simplified because all the numbers and letters inside the radical are prime factors with a power less than the index of the radical.

7 0

3 years ago

Read 2 more answers

Paha777 [63]

Step-by-step explanation:

30 minutes x an hour and half past prolly around 1 hr and 30 sec

5 0

4 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

3 years ago

HELP PLEASE 12 POINTS

Vaselesa [24]

Answer:

b. 4m

Step-by-step explanation:

congruent polygons means same shape, size, angles, and sides.

8 0

3 years ago