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borishaifa [10]

3 years ago

9

Inclues all numbers between -1 and 4 including -1 but not including 4

1 answer:

zlopas [31]3 years ago

8 0

I’m so sorry but I really don’t know the answer but try the internet it might help

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905,238 In a word form

AlladinOne [14]

Answer:

nine hundred five thousand two hundred thirty-eight

6 0

3 years ago

The dimensions of a pet carrier is 20 inches

Sergeu [11.5K]

Answer:

2702.5 inches cubed

Step-by-step explanation:

$20$ x $11\frac{3}{4}$ x $11\frac{1}{2}$ = 2702.5

3 0

2 years ago

Drag a statement or reason to each box to complete the proof.

astra-53 [7]

Answer:

The answers of the proof are the following:

2. Division Property of Equality

3. x+ 1 = -4

4. Subraction Property of Equality

5. x = -5

5 0

3 years ago

withdraws $ from bank account once each day for days. What integer represents the change in the amount in the account? Use pen

REY [17]

Answer:

a. Sam withdraws $11 from his account. That means his account balance reduces by $11 so the integer is -$11.

He does this 4 times so;

= 4 * -11

= -$44

b. He then deposits $11 once every day for 4 days.

= 4 * 11

= $44

c. The integer for withdrawals is a negative figure to show that the balance was decreasing. The Integer for deposits is positive to show that the balance was increasing.

4 0

3 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago