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3 years ago

Directed line segment AB has endpoints A(2,3) and B(8,15). Point P has coordinated P(6,11). What ratio does P partition AB into?

Mathematics

1 answer:

Vladimir79 [104]3 years ago

7 0

Refer to the graph shown below. It confirms that points A, B and P are
collinear.

Calculate the length of AP (as a).
a = √[(6 - 2)² + (11 - 3)²]
= 8.9443

Calculate the length of BP (as b).
b = √[(8 - 6)² + (15 - 11)²]
= 4.4721

Calculate ratio a/b.
a/b = 8.9443/4.4721 = 2
Therefore P partitions AB in a 2:1 ratio so that AP = 2*BP.

Answer: 2:1 ratio.

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In the accompanying diagram of circle O, chords AB and CD intersect at E and

RoseWind [281]

Answer:

m(arc AC) = 72°

m(arc BD) = 108°

m∠DEB = 90°

Step-by-step explanation:

Ratio of the measure of arc AC, arc BC, arc BD and arc AD = 4 : 2 : 6 : 8

Since, m(arc AC) + m(arc CB) + m(arc BD) + m(arc AD) = 360°

By the property of ratio,

Measure of arc AC = $\frac{4}{4+2+6+8}\times (360^0)$

= $\frac{4\times 360}{20}$

= 72°

Measure of arc BD = $\frac{6}{4+2+6+8}\times (360^0)$

= $\frac{6\times 360}{20}$

= 108°

Measure of ∠DEB = $\frac{1}{2}m(arcAC + arc BD)$

= $\frac{1}{2}(72+108)$

= 90°

6 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:

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

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

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3 years ago

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Answer:

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