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4 years ago

PLEASE HELP ME A B C OR D

Mathematics

2 answers:

RoseWind [281]4 years ago

7 0

Answer:

Examine the mathematical expression you're working on. For example, suppose you're given 5x + 3. There are two terms here, 5x and 3. Look for the variable. In this case, x is the variable. In your case its c

Find the numerical coefficient of x. Remember to look for the number before the variable. In this case, 5 is the numerical coefficient. The term 3 is a constant and is separated from the variable by a plus sign.

Keep an eye out for negative coefficients. For example, suppose you're given the statement -y + 7 + 98. You would first identify the variable, y.

Look for the numerical coefficient of y. In this case, a "1" is implied before the variable, but it's negative in this instance. Therefore, the coefficient of y is negative 1.

True [87]4 years ago

3 0

Answer:

The coefficient is D, -9.

Step-by-step explanation:

A Coefficient is the number that comes before the variable. So like, in y=6x, the coefficient would be 6. Thing is, coefficients can be either negative or positive, so you can't forget to include that. The negative really does effect the value of a number. Another example would be y=-42x. The coefficient would be -42 in that case.

Aka, it's the number before the variable.

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A two-digit locker combination is made up of nonzero digits and no digit is repeated in any combination.

frez [133]

$|\Omega|=9\cdot8=72\\
A\cap B=\{(1,2),(1,4),(1,6),(1,8)\}\\
|A\cap B|=4\\\\
P(A\cap B)=\dfrac{4}{72}=\dfrac{1}{18}\Rightarrow \text{A}$

3 0

3 years ago

What is the total resistance of a parallel circuit that has three loads? Load one has a resistance of 6 ohms. Load two has a res

gizmo_the_mogwai [7]

Answer:

The total resistance of these three resistors connected in parallel is $1.7143\Omega$

Step-by-step explanation:

The attached image has the circuit for finding the total resistance. The circuit is composed by a voltage source and three resistors connected in parallel: $R_1=6\Omega$ , $R_2=3\Omega$ and $R_3=12\Omega$ .

<u>First step: to find the source current</u>

The current that the source provides is the sum of the current that each resistor consumes. Keep in mind that the voltage is the same for the three resistors ( $R_1$ , $R_2$ and $R_3$ ).

$I_{R_1}=\frac{V_S}{R_1}$

$I_{R_2}=\frac{V_S}{R_2}$

$I_{R_3}=\frac{V_S}{R_3}$

The total current is:

$I_S=I_{R_1}+I_{R_2}+I_{R_3}=\frac{V_S}{R_1}+\frac{V_S}{R_2}+\frac{V_S}{R_3}=\frac{R_2\cdot R_3 \cdot V_S+R_1\cdot R_3 \cdot V_S+R_1\cdot R_2 \cdot V_S}{R_1\cdot R_2\cdot R_3}$

$I_S=V_S\cdot \frac{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}{R_1\cdot R_2\cdot R_3}$

The total resistance ( $R_T$ ) is the source voltage divided by the source current:

$R_T=\frac{V_S}{I_S}$

Now, replace $I_S$ by the previous expression and the total resistance would be:

$R_T=\frac{V_S}{V_S\cdot \frac{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}{R_1\cdot R_2\cdot R_3}}$

Simplify the expression and you must get:

$R_T=\frac{R_1\cdot R_2\cdot R_3}{R_2\cdot R_3+R_1\cdot R_3+R_1\cdot R_2}$

The last step is to replace the values of the resistors:

$R_T=\frac{(6\Omega )\cdot (3\Omega)\cdot (12\Omega)}{(3\Omega)\cdot (12\Omega)+(6\Omega)\cdot (12\Omega)+(6\Omega)\cdot (3\Omega)}=\frac{12}{7}\Omega=1.7143\Omega$