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3 years ago

Find the missing measure angle

Mathematics

2 answers:

azamat3 years ago

7 0

Answer:

${x}^{o} = {102}^{o}$

Step-by-step explanation:

Hint:

sum of angles = 360°

Simplify:

${x}^{o} + {78}^{o} + {90}^{o} + {90}^{o} = {360}^{o} \\ {x }^{o} + 258^{o} = {360}^{o} \\ {x}^{o} = {360}^{o} - {258}^{o} \\ {x}^{o} = {102}^{o}$

<h3>Hope it is helpful....</h3>

Goshia [24]3 years ago

5 0

Answer:

Step-by-step explanation:

x+90+90+78=360(sum of 4 angles is always 360 degree)

x+258=360

x=360-258

x=102

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What is the inverse of f(x)=x^3-2

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Would it be X=Y^3-2 bc I think you flip the x and y

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4 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:

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

Thus, the total resistance of these three resistors connected in parallel is $1.7143\Omega$

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Answer with explanation:

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