REWRITE EACH EQUATION TO ISOLATE THE INDICATED VARIABLE

Item 1

Ann [662]

Answer:

look at the attachment

Step-by-step explanation:

hope it helps!

6 0

2 years ago

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Liquid A and liquid B are mixed to make liquid C.

natta225 [31]

Answer:

Density is defined as:

Density = mass/volume.

We know that:

For liquid A:

Density = 70kg/m^3

Mass = 1400kg

Then the volume is:

Volume = mass/density = (1400kg)/(70kg/m^3) = 20 m^3

For liquid B:

Density = 280 kg/m^3

Volume = 30m^3

We can find the mass of liquid B as:

mass = density*volume = (280kg/m^3)*(30m^3) = 8400 kg

We know that liquid C is a mixture of liquid A and B.

Then the mass of liquid C will be equal to the sum of the masses of liquid A and B, then:

Mass of liquid C = 1400kg + 8400kg = 9800kg

The same happens for the volume, then:

Volume of liquid C = 30m^3 + 20m^3 = 50m^3

Then the density of liquid C is:

Density of liquid C = (9800kg)/(50m^3) = 196 kg/m^3

6 0

3 years ago

Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]

alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

$\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]$

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)² = 3√6

Therefore, the unit vector is

$\frac{[7,2,-1]}{3\sqrt{6} }$

or

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

In conclusion, the two unit vectors are;

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

and

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

<em>Hope this helps!</em>

7 0

2 years ago

Select the correct answer.<br> Which graph shows a function and its inverse?

Anit [1.1K]

Answer:

first graph shows the function and it's inverse

8 0

2 years ago

Graph of a cubic polynomial that falls to the left and rises to the right with x intercepts negative 2, negative 1, and 2

xz_007 [3.2K]

Hello,

A cubic polynomial with -2,1,2 as roots is
f(x)=k*(x+2)(x+1)*(x-2)=k(x²-4)(x+1)=k(x^3+x²-4x-4)

Looking the answers ==>k=1

==>Answer A

6 0

3 years ago

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