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3 years ago

8

O support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground. Th

e tree meets the ground at a right angle.
At approximately what angle does the wire meet the ground?

33.6°
39.8°
50.2°
56.4°

2 answers:

lora16 [44]3 years ago

6 0

The answer would be 56.4° also you’re welcome

Viktor [21]3 years ago

4 0

The answer could be 56.4

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Find the cardinal number of the set<br> C = {x|x € N, 30 < x < 56 and x is divisible by 6},

Naddik [55]

<h3>Answer: 4</h3>

=====================================================

Explanation:

We're given the following

$C = \{ x | x \in \mathbb{N}, \ 30 < x < 56, \ \text{ x is divisible by 6}\}$

which means that we have a set of natural numbers between 30 and 56, excluding both endpoints, and the set of numbers are also multiples of 6.

So looking at the multiples of 6 that are between 30 and 56 leads us to

{30, 36, 42, 48, 54}

but we cross off 30 as we want x to be larger than that, so we get this updated final list

{36, 42, 48, 54}

There are four items in this final list so the cardinality of the set described is 4. The cardinality is just the number of items in a given set.

4 0

4 years ago

Using the figure below, find the measure of z ( just enter in the number for your answer don’t worry about the degree symbol or

belka [17]

Answer:

z = 29°

Step-by-step explanation:

We know that the 3 angles in a triangle always add up to 180°. So for triangle ABC, angle A = 59° (given), and angle B = 43° + 49° = 92° (given).

angle A + angle B + angle C = 180°

59° + 92° + z = 180°

151° + z = 180°

Subtract 151° from both sides of the equation

z = 180° -151°

z = 29°

8 0

3 years ago

A force of 7 lb is required to hold a spring stretched 9 in. beyond its natural length. How much work W is done in stretching it

Ymorist [56]

144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Force F=10 lb

From the hookes Law

$F=kx$

Therefore, calculate k for the spring for

$10=k(\frac{4}{12}) = > k=\frac{120}{4}=30$

Work done in stretching a spring through a length dx is

$dW=F.dx = > dW=kxdx$

For calculating the work done for stretching the spring to x=6in=(6/12)feet beyond the natural length, Integrate over the limits of x=0 to x=1/2

Therefore,

$\int_{0}^{W}dW=\int_{0}^{0.5}kxdx = > W=\frac{1}{2}k[x^2]_{0}^{0.5}=\frac{1}{2}30 \times (0.5)^2=\frac{30}{8}$

That is,

$W=\frac{15}{4} ft-lb\\$

The answer that is given is for stretching the spring to 4 inches.

Mass of 10 m of chain length is 80kg. This implies, Mass per unit meter length of the chain is

$m_{l}=\frac{80}{10}=8kg/m$

Consider a small length dx of the chain at the end point A. Work done in lifting the small length dx over the height of x meters is

$dW=(8g)dx \times x=8gxdx$

Now, integrate over the the value of x from

x=0 when end of the chain A is on the ground

x=6 when end of the chain A reaches 6 m above the ground

That is,

$\int_{0}^{W}dW=\int_{0}^{6}8gxdx=8g [\frac{x^2}{2}]_{0}^{6}=4g(6^2-0^2)$

$W=4g \times 36=144g \ Joules$

Hence,144 joules is the work W done in stretching it from its natural length to 13 in beyond its natural length

Learn more about Integration here brainly.com/question/2263647

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7 0

2 years ago

Determine the midpoint between the points (-5, 3) and (5, 9).

PSYCHO15rus [73]

Answer:

M (0, 6)

Step-by-step explanation:

By midpoint formula,

M [(x1+x2)/2 , (y1+y2)/2]

M [-5+5/2 , 3+9/2]

M (0/2 , 12/2)

M (0, 6)

4 0

2 years ago

Your mom asked you to turn the oven down to 325 degrees. This is 75 degrees less than it was. What was the original temperature?

Paladinen [302]

I'm assuming 325 + 75 = 400.

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3 years ago