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3 years ago

Answer the question I need help

Mathematics

2 answers:

Inessa05 [86]3 years ago

6 0

Answer:

the answer would be 38 :)

Step-by-step explanation:

24÷3=8

(8)+3+3 to the 3rd power

3×3×3=27

8+3=12 ->12+27=38

hopefully this helps :)

swat323 years ago

4 0

Answer:

Step-by-step explanation:

step 1: 24/3 which equals to 8.

Step 2: 8+3= 11

Step 3: 3*3*3=27

step four: 11+27=20 answer=38

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RESUELVE EL SIGUINTE PORBLEMA En la escuela "Soy Feliz porque soy diverso", hay 230 estudiantes en total. De estos, 35 son mujer

Tanzania [10]

Answer:

1) a. 195 are males

b. Males are the 85 %

2) 1/2

Step-by-step explanation:

Number of all the students: 230

35 are females

x are males

Then 35 + x = 230

x = 230 - 35 → 195

To determine the amount of males, from the total we apply the percent.

Percent = (Amount of males / Total of students) . 100

% = (195 / 230) . 100 = 85 %

2. In first place, Maria spent 13 for the book.

In second place, Maria spent 29 for the pencil box.

In total, she has spent 13+29 = 42

But, how many dolars have the grandpa gave to her?.

We can also apply percent and fractions, where:

(Book / Total) + (Pencil box/Total) + (What she did not spend/Total) = 1

13 + 29 + x = T

13/T + 29/T + x/T = 1, in conclussion x = 42

Total of money given = 84

Then, Maria has not spent the half of money→ 1/2

7 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 $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .