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Elanso [62]
3 years ago
14

Shannon likes to jog around a lake

Mathematics
1 answer:
just olya [345]3 years ago
6 0

Shannon jogs 1 mile

440 x 4 = 1,760

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How many solutions does the following equation have?
dedylja [7]
B. Exactly one solution

7y +21 = 5y +8
7y +13=5y
13=-2y
-6.5=y
3 0
3 years ago
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Can someone help? Show work
Kamila [148]
Given the isosceles right triangle in which each of its legs has a measure of 4:

Let c = hypotenuse
a = one of the legs of the triangle
b = the other leg of the triangle

Use the Pythagorean Theorem to solve for the measure of the hypotenuse:

c^ 2 = a^2 + b^2

c^ 2 = (4)^2 + (4)^2
c^ 2 = 16 + 16
c^2 = 32

Next, take the square root of each side of the equation to solve for c:

√(c^2) = √(32)

c = 4√(2)

Therefore, the correct answer is Option 3) 4√(2).
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3 years ago
The square pyramid pictured below has a surface area of<br> 9 m<br> 6 m<br> 6 m
Genrish500 [490]
Hey,just to let you know that they is no picture as I can’t answer it.
5 0
3 years ago
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in a farm there are 600 mango trees. the number of trees in each row is 10 less than twice the number of rows. does each row hav
iren [92.7K]

The area of the farm is the number of trees on it.

Each row of the farm does not have 50 trees

Let

<em />x \to<em> number of rows</em>

<em />y \to<em> number of trees in each row</em>

<em />

So, we have:

<em />y = 2x - 10<em> --- twice less than the number of rows</em>

The area of the farm is:

Area = x \times y

Where

Area = 600 --- the number of trees

So, we have:

x \times y = 600

Substitute y = 2x - 10

x \times (2x - 10) = 600

Substitute 50 for x

50 \times (2 \times 50 - 10) = 600

50 \times (100 - 10) = 600

50 \times 90 = 600

4500 = 600

The above equation is not true.

This means that, each row does not have 50 trees

Read more about areas at:

brainly.com/question/16418397

4 0
2 years ago
How to find the length of a triangle with only one side non right triangle?
castortr0y [4]
The trigonometry of non-right triangles

So far, we've only dealt with right triangles, but trigonometry can be easily applied to non-right triangles because any non-right triangle can be divided by an altitude* into two right triangles.

Roll over the triangle to see what that means →



Remember that an altitude is a line segment that has one endpoint at a vertex of a triangle intersects the opposite side at a right angle. See triangles.

Customary labeling of non-right triangles

This labeling scheme is comßmonly used for non-right triangles. Capital letters are anglesand the corresponding lower-case letters go with the side opposite the angle: side a (with length of a units) is across from angle A (with a measure of A degrees or radians), and so on.



Derivation of the law of sines

Consider the triangle below. if we find the sines of angle A and angle C using their corresponding right triangles, we notice that they both contain the altitude, x.



The sine equations are



We can rearrange those by solving each for x(multiply by c on both sides of the left equation, and by a on both sides of the right):



Now the transitive property says that if both c·sin(A) and a·sin(C) are equal to x, then they must be equal to each other:



We usually divide both sides by ac to get the easy-to-remember expression of the law of sines:



We could do the same derivation with the other two altitudes, drawn from angles A and C to come up with similar relations for the other angle pairs. We call these together the law of sines. It's in the green box below.

The law of sines can be used to find the measure of an angle or a side of a non-right triangle if we know:

two sides and an angle not between them ortwo angles and a side not between them.

Law of Sines



Examples: Law of sines

Use the law of sines to find the missing measurements of the triangles in these examples. In the first, two angles and a side are known. In the second two sides and an angle. Notice that we need to know at least one angle-opposite side pair for the Law of Sines to work.

Example 1

Find all of the missing measurements of this triangle:




The missing angle is easy, it's just



Now set up one of the law of sines proportions and solve for the missing piece, in this case the length of the lower side:



Then do the same for the other missing side. It's best to use the original known angle and side so that round-off errors or mistakes don't add up.



Example 2

Find all of the missing measurements of this triangle:




First, set up one law of sines proportion. This time we'll be solving for a missing angle, so we'll have to calculate an inverse sine:



Now it's easy to calculate the third angle:



Then apply the law of sines again for the missing side. We have two choices, we can solve



Either gives the same answer,



Derivation of the law of cosines

Consider another non-right triangle, labeled as shown with side lengths x and y. We can derive a useful law containing only the cosine function.



First use the Pythagorean theorem to derive two equations for each of the right triangles:



Notice that each contains and x2, so we can eliminate x2 between the two using the transitive property:



Then expand the binomial (b - y)2 to get the equation below, and note that the y2 cancel:



Now we still have a y hanging around, but we can get rid of it using the cosine solution, notice that



Substituting c·cos(A) for y, we get



which is the law of cosines

The law of cosines can be used to find the measure of an angle or a side of a non-right triangle if we know:

two sides and the angle between them orthree sides and no angles.

We could again do the same derivation using the other two altitudes of our triangle, to yield three versions of the law of cosines for any triangle. They are listed in the box below.

Law of Cosines

The Law of Cosines is just the Pythagorean relationship with a correction factor, e.g. -2bc·cos(A), to account for the fact that the triangle is not a right triangle. We can write three versions of the LOC, one for every angle/opposite side pair:



Examples: Law of cosines

Use the law of cosines to find the missing measurements of the triangles in these two examples. In the first, the measures of two sides and the included angle (the angle between them) are known. In the second, three sides are known.


3 0
3 years ago
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