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2 years ago

Max's house was 4 kilometers from Sam's house.how many meters apart are the two houses

Mathematics

1 answer:

Fofino [41]2 years ago

7 0

4000
kilo means 1000
so you gust have to multiply 4 by a thousand

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What value of z makes the statement true?

Shkiper50 [21]

-(z + (-17)) = 12 + 17

-(z - 17) = 12 + 17

Distribute -1:

-z + 17 = 12 + 17

Subtract 17 to both sides:

-z = 12

Multiply -1 to both sides:

z = -12

4 0

3 years ago

Two dice are rolled. E is the event that the sum is even, F is the event of rolling at least one six, and G is the event that th

mrs_skeptik [129]

Answer:

(a) $E \cap F ={(2, 6),(4, 6),(6, 2),(6, 4),(6, 6)}$

(b) $E^c \cap G =\{ \}$

Step-by-step explanation:

The sample space of two dice rolled is given below:

$\{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)\\(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)\\(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\\(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\\(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),(5,6)\\(6, 1), (6, 2), (6, 3), (6, 4), (6,5), (6,6)\}$

For Event E (The sum is even), the outcomes are:

$(1, 1), (1, 3), (1, 5),(2, 2), (2, 4), (2, 6)\\(3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6)\\(5, 1),(5, 3), (5, 5), (6, 2), (6, 4), (6,6)$

For Event F (Rolling at least one six), the outcomes are:

$(1, 6), (2, 6), (3, 6), (4, 6),(5,6),(6, 1), (6, 2), (6, 3), (6, 4), (6,5), (6,6)$

For Event G (The sum is eight), the outcomes are:

$(2, 6), (3, 5),(4, 4), (5, 3),(6, 2)$

(a) $E \cap F$

Therefore:

$E \cap F ={(2, 6),(4, 6),(6, 2),(6, 4),(6, 6)}$

(b) $E^c \cap G$

E is the event that the sum is even

Therefore: $E^c$ is the event that the sum is odd.$

Since G is the event that the sum is eight( which is even), the intersection of the complement of E and G will be empty.

Therefore:

$E^c \cap G =\{ \}$

5 0

2 years ago

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. Below, enter your answers so that ∠A

wariber [46]

Answer:

$\angle B_{1} \approx 31.668^{\circ}$ , $\angle B_{2} \approx 148.332^{\circ}$

$\angle C_{1} \approx 127.332^{\circ}$ , $\angle C_{2} \approx 10.668^{\circ}$

$c_{1} \approx 157.532$ , $c_{2}\approx 36.676$

Step-by-step explanation:

The Law of Sines states that:

$\frac{a}{\sin A} = \frac{b}{\sin B}=\frac{c}{\sin C}$

Where:

$a$ , $b$ , $c$ - Side lengths, dimensionless.

$A$ , $B$ , $C$ - Angles opposite to respective sides, dimensionless.

Given that $a = 71$ , $b = 104$ , $\angle A = 21^{\circ}$ , the sine of angle B is:

$\sin B = \frac{b}{a}\cdot \sin A$

$\sin B = \frac{104}{71}\cdot \sin 21^{\circ}$

$\sin B = 0.525$

Sine is positive between 0º and 180º, so there are two possible solutions:

$\angle B_{1} \approx 31.668^{\circ}$

$\angle B_{2} \approx 148.332^{\circ}$

The remaining angle is obtained from the principle that sum of internal triangles equals to 180 degrees: ( $\angle A = 21^{\circ}$ , $\angle B_{1} \approx 31.668^{\circ}$ , $\angle B_{2} \approx 148.332^{\circ}$ )

$\angle C_{1} = 180^{\circ}-\angle A - \angle B_{1}$

$\angle C_{1} = 180^{\circ}-21^{\circ}-31.668^{\circ}$

$\angle C_{1} \approx 127.332^{\circ}$

$\angle C_{2} = 180^{\circ}-\angle A - \angle B_{2}$

$\angle C_{2} = 180^{\circ}-21^{\circ}-148.332^{\circ}$

$\angle C_{2} \approx 10.668^{\circ}$

Lastly, the remaining side of the triangle is found by means of the Law of Sine: ( $a = 71$ , $\angle A = 21^{\circ}$ , $\angle C_{1} \approx 127.332^{\circ}$ , $\angle C_{2} \approx 10.668^{\circ}$ )

$c_{1} = a\cdot \left(\frac{\sin C_{1}}{\sin A} \right)$