Ask question

Ask question

All categories

3 years ago

13

Definition of inequality

2 answers:

fenix001 [56]3 years ago

7 0

What is it

Jejejfjfjjdjfjfkfj

Rus_ich [418]3 years ago

6 0

DescriptionIn mathematics, an inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expressions. It is used most often to compare two numbers on the number line by their size.

You might be interested in

Find f(6) if f(x) = x^{2} ÷ 3 + x.

Papessa [141]

6^2 is 36. / 3 makes 12. 12+6 is 18. final answer, most likely 18.

8 0

3 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

Consider the circle with a radius of 30 cm. What is the approximate circumference of the circle? (use 3.14 as an approximation o

Goshia [24]

Answer: 188.4 cm

Step-by-step explanation:

Given

the radius of the circle is $r=30\ cm$

Also, circumference is given by $P=2\pi r$

Putting values

$\Rightarrow P=2\times 3.14 \times 30=188.4\ cm$

The circumference of the circle is $188.4\ cm$

8 0

3 years ago

The half-life of a radioactive substance is the time required for half of a sample to undergo radioactive decay, or for the quan

Ilya [14]

That's two half lives. 5/8 * 2 * 2 = 20/8 = 5/2 pounds7/8 * 2 * 2 = 28/8 = 7/2 pounds

6 0

3 years ago

Read 2 more answers

Simplify to create an equivalent expression.<br> 2-4(5p+1)}2−4(5p+1)

ZanzabumX [31]

Answer:

-8-60p

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers