How can you represent the system of inequalities with a matrix? 11x-12y=7 5x+12y=-7

Diamonds have a density of 3.5 LaTeX: \frac{g}{cm^3}g c m 3. How big is a diamond that has a mass of 0.10 g?

liubo4ka [24]

For this case we have that by definition, the density is given by:

$d = \frac {M} {V}$

Where:

M: It is the mass of the diamond

V: It is the volume of the diamond

According to the data of the statement we have:

$d = 3.5 \frac {g} {cm ^ 3}\\M = 0.10 \ g$

So the volume is:

$V = \frac {M} {d}\\V = \frac {0.10 \ g} {3.5 \frac {g} {cm ^ 3}}\\V = 0.02857\\V = 0.03 \ cm^3$

Thus, the volume of the diamond is approximately $0.03 \ cm ^ 3$

Answer:

$0.03 \ cm ^ 3$

3 0

3 years ago

Factor the expression completely over the complex numbers. y^4+12y^2+36

Brums [2.3K]

Y⁴ + 12y² + 36
Now factorize the expression
y⁴ + 6y² + 6y² + 36
= y²(y² + 6) + 6(y² + 6)
= (y² + 6) (y² + 6)
Now 6 is not the perfect square and according to rule, binomial can not be factored as the difference of two perfect squares.
so multiply both.
(y² + 6)² is the answer.

7 0

2 years ago

Read 2 more answers

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

FInd the value of x. Y D W X (15x-8) X = N 226

Oksi-84 [34.3K]

Let's start with the given arc and its angle.

The angle YWX is going to be half the arc length.

YWX = 1/2 (226) = 113

Angles VWX and YWX form a linear pair (or are supplementary angles), which means that their sum is 180 degrees.

(15x - 8) + 113 = 180

15x + 105 = 180

15x = 75

x = 5

Hope this helps!

8 0

2 years ago

Gross worldwide sales in Mattel's Wheels (Hot Wheels) brand was $885.1 million in 2016 and $847 million in 2017. Calculate the a

attashe74 [19]

Answer:

Step-by-step explanation:

Absolute Change = final change in 2017 sales - initial change in 2016 sales = $847,000,000 - $885,100,000 = 38,100,000

Relative Change = (Final - Initial change)/initial change x 100% = 38,100,000/885,100,00 x 100% = 381/8851 x 100% = 0.0430459835 x 100% = 4.30%

6 0

2 years ago