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3 years ago

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How many liters are equal to 5,890 milliters? Use the placement of the decimal point when dividing by a power of 10 to help you

sol
5.89 L
58,900 L
58.9 L
589 L

1 answer:

Oliga [24]3 years ago

5 0

Answer:

5.89 liters

Step-by-step explanation:

1 liter is equal to 1000 mL

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If m∠A = 72°, m∠B = 32°, and c = 8, what are the measures of the remaining sides and angle?

mixas84 [53]

The laws of cosine and sine and the given parameters can be used to

calculate the measure of angles and distances.

Correct responses:

Question 1: A. m∠C = 76°, a = 7.84, b = 4.37

Question 2: C. 1 triangle

Question 3: D. 31.2 miles

Question 4: A. b = 10.799

Question 5: C. 28.9°

<h3>Methods and calculations used to obtain the above responses </h3>

Question 1:

If m∠A = 72°

m∠B = 32°

c = 8

Therefore;

m∠C = 180° - 72° - 32° = 76°

$\displaystyle \frac{a}{sin(72^{\circ})} = \mathbf{ \frac{8}{sin(76^{\circ})} }$

$\displaystyle a = \mathbf{\frac{8}{sin(76^{\circ})} \times sin(72^{\circ})} \approx 7.84$

$\displaystyle b = \frac{8}{sin(76^{\circ})} \times sin(32^{\circ}) \approx 4.37$

The correct option is therefore;

<u>A. m∠C = 76°, a ≈ 7.84, b ≈ 4.37</u>

Question 2

The given parameters are;

$\displaystyle B = \mathbf{ \frac{\pi}{6} }$

$\displaystyle \frac{\pi}{6} = 30^{\circ}$

a = 20

b = 10

Therefore, by the law of cosine we have;

b² = a² + c² - 2·a·c·cos(B)

Which gives;

10² = 20² + c² - 2×20×c×cos(30°)

100 = 400 + c² - 20·c·(√3)

c² - 20·√3 ·c + 300 = 0

$\displaystyle c = \frac{20 \cdot \sqrt{3} \pm\sqrt{1200-4 \times 1 \times 300} }{2 \times 1} = \mathbf{ 10 \cdot \sqrt{3} }$

Therefore, given that <em>c</em> has only one value, the three sides of the triangle are known, and the number of triangles unique triangles by Side-Side-Side description is; <u>C. 1 triangle</u>

Question 3

The distance between the weather station = 24 miles

The bearing of the storm from weather station A = N17°W

The bearing of the storm from weather station B = N48°W

The angle formed at point <em>C</em> in ΔABC is therefore;

108° - 42° - 107° = 31°

By sine rule, we have;

$\displaystyle \frac{24}{sin(31^{\circ})} =\mathbf{\frac{x}{sin(42^{\circ})} }$

Where:

x = The distance from weather station <em>A</em> from the storm

Which gives;

$\displaystyle x = \frac{24}{sin(31^{\circ})} \times sin(42^{\circ}) \approx \mathbf{31.2 \ miles}$

The distance from weather station <em>A</em> from the storm is <u>D. 31.2 miles</u>

Question 4

∠A = 52°

∠C = 57°

Side BC = 9

Therefore;

∠B = 180° - 52° - 57° = 71°

∠B = 71°

According to the law of sines, we have;

$\displaystyle \frac{b}{sin(71^{\circ})}= \mathbf{ \frac{9}{sin(52^{\circ})} }$

Therefore;

$\displaystyle b = \frac{9}{sin(52^{\circ})} \times sin(71^{\circ}) \approx \mathbf{ 10.799}$

The correct option is; <u>A. b = 10.799</u>

Question 5

Given:

Label of the vertex of the triangle formed are; The golfer, spectator, hole

Distance from the golfer to the hole = 200 yards

Distance from the golfer to the spectator = 140 yards

Vertex angle at the spectator = 110°

By using the law of sines, we have;

$\displaystyle \frac{200}{sin(110^{\circ})} = \frac{140}{sine \ of \ the \ angle \ at \ the \ hole, \ \phi}$

Therefore;

$\displaystyle sin(\phi) = \mathbf{ \frac{140}{200} \times sin(110^{\circ})} = 0.7 \times sin(110^{\circ})$

The angle at the hole, ∅ = arcsin(0.7×sin(110°)) ≈ 41.13°

Therefore;

The angle that the golfer has = 180° - 110° - 41.1° = 28.9°

The angle that the golfer has between the spectator and the hole is <u>C. 28.9°</u>

Learn more about law of cosine and sine here:

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Find the sum of 3+1/2

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Answer:

3/2 or 1 1/2

Step-by-step explanation:

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3 years ago

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Find the slope of (3,43) and (8,113)

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Answer: Y=14x+1

When x=0 Y=1

When Y=0, X= 0.071428571428571

Hope this helps. :)

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3 years ago

given the weekly demand curve of a local wine producer is p= 50-0.1q, and that the total cost function is c= 1500+ 10q, where q

ipn [44]

Answer:

a)the weekly profit as a function of price is $P=-10 p^2 + 600 p - 6500$

b) a bottle of wine be sold at $30 to realise maximum profit

c) the maximum profit that can be made by the producer is $2500

Step-by-step explanation:

The weekly demand curve of a local wine producer is p= 50-0.1q

p = price

q = quantity

Revenue function R =Price \times Quantity

R=(50-0.1q)q

$R=50q-0.1q^2$

Cost function: c= 1500+ 10q

Profit function=R(x)-C(x)

Profit function= $50q-0.1q^2-1500-10q$

Profit function= $40q-0.1q^2-1500$ ----1

We have q = 500 - 10 p using p = 50 − 0.1q

$P=-0.1 (500 - 10 p)^2 + 40 (500 - 10 p)- 1500\\P= -10 p^2 + 600 p - 6500$

General quadratic equation: $ax^2+bx+c=0$

On comparing

a = -0.1 , b = 40 , c = -1500

Maximum Profit is at q = $\frac{-b}{2a}=\frac{-40}{2(-0.1)}=200$

To find price must a bottle of wine be sold to realise maximum profit

p= 50-0.1q

p= 50-0.1(200)=30

Substitute the value of q in profit function(1) get the maximum profit

So, Profit function= $40\left(200\right)-0.1\left(200\right)^{2}-1500=2500$

Hence

a)the weekly profit as a function of price is $P=-10 p^2 + 600 p - 6500$

b) a bottle of wine be sold at $30 to realise maximum profit

c) the maximum profit that can be made by the producer is $2500

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What’s the square root of 6

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hey friend, the root of 6 is 2.44

Step-by-step explanation:

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