All categories

4 years ago

How do you know that 21 over 30 is greater than 2 over 3

Mathematics

2 answers:

fgiga [73]4 years ago

7 0

I know that 21 over 30 is greater than 2 over 3 because 3 is bigger that 2 and the 2 is less than

NISA [10]4 years ago

6 0

21 over 30 is greater than 2 over 3 because 2 over 3 times 10 is 20 over 30. And 21 over 30 is greater than 21 over 30.

You might be interested in

(5t^3)^-4 1. 625/t^12 2. 20/t^7 3. 1/625t^12 4. 1/20t^7

rusak2 [61]

Answer:

$\large\boxed{3.\ \dfrac{1}{625t^{12}}}$

Step-by-step explanation:

$\text{Use}\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\---------------\\\\(5t^3)^{-4}=\dfrac{1}{(5t^3)^4}=\dfrac{1}{5^4(t^3)^4}=\dfrac{1}{625t^{(3)(4)}}=\dfrac{1}{625t^{12}}$

4 0

3 years ago

The length of a rectangle is (x-8) units, and its width is(x+11) units. What is an expression for this.

Ipatiy [6.2K]

Answer:

(x²+3x-88) square units

Step-by-step explanation:

Area of a rectangle = Length × Width

Given

length of a rectangle is (x-8)units

Its width is (x+11) units.

Required

Expression for the area of the rectangle

Substituting the given function into the formula.

Area of the rectangle = (x-8)(x+11)

Area of the rectangle = x(x)+11x-8x-88

Area of the rectangle = x²+3x-88

Hence the expression that represents the area of the rectangle is x²+3x-88 units²

8 0

3 years ago

Given that the quadrilateral EFGH is a trapezoid, m

I am Lyosha [343]

Answer:

I need more information

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

PLEASE HELP!!

irina [24]

Answer:

$\leadsto make \: y \: subject \\ - 3y + 2x = 5y - 8 \\ - 3y - 5y = - 2x - 8 \\ y( - 3 - 5) = - 2x - 8 \\ - 8y = - 2x - 8$

divide through out by -8 :

${ \boxed{ \boxed{y = \frac{1}{4} x + 1}}}$

Answer is b

3 0

3 years ago

A box with a square base and open top must have a volume of 157216 cm3. We wish to find the dimensions of the box that minimize

shepuryov [24]

Answer:

Base Length of 68cm
Height of 34 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 157216 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume $=x^2h=157216$

$h=\dfrac{157216}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{157216}{x^2}\\A(x)=x^2+4x\left(\dfrac{157216}{x^2}\right)\\A(x)=\dfrac{x^3+628864}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+628864}{x}\\A'(x)=\dfrac{2x^3-628864}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-628864}{x^2}=0\\2x^3-628864=0\\2x^3=628864\\x^3=314432\\x=\sqrt[3]{314432}\\ x=68$

Step 4: Verify that x=68 is a minimum value

We use the second derivative test

$If\:A(x)=\dfrac{x^3+628864}{x}\\A''(x)=\dfrac{2x^3+1257728}{x^3}\\$When x=68\\A''(x)=6$

Since the second derivative is positive at x=68, then it is a minimum point.

Recall:

$h=\dfrac{157216}{x^2}\\h=\dfrac{157216}{68^2}=34$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 68cm
Height of 34 cm.

3 0

3 years ago