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3 years ago

Please help me with 4 and 5

Mathematics

1 answer:

Sedbober [7]3 years ago

3 0

Answer:

The answer to No. 4 is x = 12.

The answer to No. 5 is 160.

Step-by-step explanation:

(No. 4) The mean (average) is calculated by adding up all the numbers and then dividing by how many numbers there are.

In the first exercise, there are 6 number with a mean of 8.5. Since the mean is calculated by dividing the sum of the numbers by how many numbers there are, we know that the sum of the numbers have to be 8.5 × 6 = 51.

Subtracting all the known numbers from 51 leaves you with the only possible solution for x: 51 - 11 - 8 - 8 - 7 - 5 = 12.

(No. 5) Start by calculating the total mass of all the spiders. (16 × 175) + (24 × 150) = 6,400. Once again, to calculate the mean, divide the total sum of the numbers by how many numbers there are, in this case you divide the total mass of the spiders by how many spiders there are, leaving you with the average of 6,400 ÷ 40 = 160.

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3 years ago

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3 years ago

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Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

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Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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3 years ago