Which of the following answers is the correct one, explain

a pentagon is a regular pentagon that has 5 sides, so each side has the same length that is (x - 1.5)cm. The perimeter is 22.5 c

Ivanshal [37]

Answer:

Step-by-step explanation:

5(x - 1.5) = 22.5

5x - 7.5 = 22.5

5x = 15

x = 3

8 0

3 years ago

Consider the function,

RUDIKE [14]

Answer:

If x= 4 then f(x) = 4x -5 is 11.

Step-by-step explanation:

f(x) = 4x -5

We need to find the domain value that corresponds to the output f(x) = 11

In this question, we need to solve the expression for value of x such that the answer is 11.

if x= 3

f(3) = 4(3) -5

= 12 -5

= 7

Since we want the answer 11 so we cannot take x= 3

if x = 4

f(4) = 4(4)-5

= 16 - 5

= 11

So, if x= 4 then f(x) = 4x -5 is 11.

5 0

3 years ago

A toy company produces two kites whose shapes are geometrically similar. Find the length of the missing side of the smaller kite

docker41 [41]

M=30/22,5=1,33
m=25/x=1,33. ...>x=25/1,33=18,7

5 0

3 years ago

Can you help me with this please

krek1111 [17]

Answflasvkdiuihrytiy\ih\j

5 0

3 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