If you +3+5 what would be your simplify answer

Devon made a box with length x+1, width x+3, and height x-3. What is the volume of Devon's box as a function of x?

erica [24]

Answer:

(a)

$V=x^3+x^2-9x-9$

(b)

$x=10$

(c)

$x=\frac{7}{2}$

Step-by-step explanation:

we are given

length is x+1

$L=x+1$

width is x+3

$W=x+3$

height is x-3

$H=x-3$

(a)

we can use volume formula

$V=L\times W\times H$

now, we can plug it

$V=(x+1)\times (x+3)\times (x-3)$

now, we can simplify it

$V=x^3+x^2-9x-9$

(b)

we are given volume =1001

so, we can set V=1001

and then we can solve for x

$V=x^3+x^2-9x-9=1001$

now, we can factor it

$\left(x-10\right)\left(x^2+11x+101\right)=0$

$x-10=0$

$x=10$

(b)

we are given volume

so, we can set

$V=14\frac{5}{8}=\frac{14\times 8+5}{8}$

$V=\frac{117}{8}$

and then we can solve for x

$V=x^3+x^2-9x-9=\frac{117}{8}$

now, we can factor it

$8x^3+8x^2-72x-189=0$

$\left(2x-7\right)\left(4x^2+18x+27\right)=0$

$2x-7=0$

$x=\frac{7}{2}$

8 0

3 years ago

If 4 Maths books are selected from 6 different Maths different English books, how many ways can the seven om Maths books and 3 E

alexgriva [62]

Answer:

Please see the answer below

Step-by-step explanation:

a. Since there’s no restrictions .Therefore , the number of ways = 7!*150 = 756000

b. The number of ways such that the 4 math books remain together

The pattern is as follows: MMMMEEE, EMMMMEE, EEMMMME, and EEEMMMM

Where M = Math’s Book and E= English Book.

Number of ways = 4!*8!*4*150= 86400 ways.

c. The number of ways such that math book is at the beginning of the shelf

The number of ways = 6!*4*150 = 432000

d. The number of ways such that math and English books alternate

The number of ways = 150*4!*3! =2160 ways

e. The number of ways such that math is at the beginning and an English book is in the middle of the shelf. The number of ways = 4*3*5!*150 =216000 ways.

6 0

3 years ago

Factorise 2x^2 - 3x -2 < 0

Galina-37 [17]

Step 1: Factor $2 x^{2} - 3x -2$
1. Multiply 2 by -2, which is -4.
2. Ask: Which two numbers add up to -3 and multiply to -4?
3. Answer: 1 and -4
4. Rewrite $-3x$ as the sum of $x$ and $-4x$

$2 x^{2} +x-4x-2\ \textless \ 0$

Step 2: Factor out common terms in the first two terms, then in the last two terms.

 $x(2x+1)-2(2x+1)\ \textless \ 0$

Step 3: Factor out the common term $2x+1$

$(2x+1)(x-2)\ \textless \ 0$

Step 4: Solve for $x$

1. Ask: When will $(2x+1)(x-2)$ equal zero?
2. Answer: When $2x + 1 = 0$ or $x-2=0$
3. Solve each of the 2 equations above:

 $x=- \frac{1}{2} ,2$

Step 5: From the values of $x$ above, we have these 3 intervals to test.
x = < -1/2
-1/2 < x < 2
x > 2

Step 6: Pick a test point for each interval

For the interval $x\ \textless \ - \frac{1}{2} 
$
Lets pick $x=-1.$ Then, $2(-1) ^{2} -3 * -1 -2 \ \textless \ 0$
After simplifying, we get $3\ \textless \ 0$ , Which is false.
Drop this interval.

For this interval $- \frac{1}{2} \ \textless \ x\ \textless \ 2$
Lets pick $x=0$ . Then, $2* 0^{2} - 3 * 0-2\ \textless \ 0$ . After simplifying, we get $-2\ \textless \ 0,$ which is true. Keep this interval.

For the interval $x\ \textgreater \ 2$

Lets pick $x = 3.$ Then, $2 * 3 ^{2} -3*3-2\ \textless \ 0.$ After simplifying, we get $7\ \textless \ 0$ , Which is false. Drop this interval.

.Step 7: Therefore,
$- \frac{1}{2} \ \textless \ x\ \textless \ 2$

Done! :)

4 0

3 years ago

220 students were asked if they liked chocolate or vanilla ice cream. 140 said they liked chocolate and 120 said they liked vani

Flura [38]

Answer: D. 40

Step-by-step explanation:

Given: Number of students liked vanilla or chocolate ice cream = $n(V\cup C)=220$