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2 years ago

How does kidney stones affect the excretory system ?

1 answer:

8 0

<em>Kidney stones can stay in the kidney or travel down the urinary tract. A small stone may pass on its own, causing little or no pain. A larger stone may get stuck along the urinary tract. When a stone gets stuck it can block the flow of urine this causes severe pain and bleeding.</em>

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Dr. Paul found that college students between the ages of 18 and 23 performed better on a cognitive task in a low-noise condition

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3 years ago

Please help me im not understanding this...

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2 years ago

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2 years ago

1. What type of democracy developed in Athens?

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Answer:

1. direct democracy

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3 years ago

Why did the professional dog walker go out of business math worksheet answers?

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Why Did the Professional Dog Walker ao Out of business, math worksheet Answers:

Q1. sin27° = x/8

Solution:

We have to solve for x, therefore, we will rearrange the given equation for x.

We get,
x = 8 × sin27°

Using the calculator,

sin27° = 0.45

Now substitute the value of sin27° into the main equation.

we get,
x = 8 × 0.45
x = 3.63 (rounded to the nearest hundredth)

Q2. tan 18° = n / 75

Solution:
We have to solve for n, therefore, we will rearrange the given equation for n.
We get,
n = 75 × tan 18°
Using the calculator,
tan 18° = 0.32
Now substitute the value of tan 18° into the main equation.
we get,
x = 75 × 0.32
x = 24.37 (rounded to the nearest hundredth)

Q3. sin40° = 4 / a

Solution: We have to solve for a, therefore, we will rearrange the given equation for a.
We get,
a = 4 ÷ sin40°
Using the calculator,
sin40° = 0.64
Now substitute the value of sin40° into the main equation.
we get,
a = 4 ÷ 0.64
a = 6.25 (rounded to the nearest hundredth)

Q4. cos5° = 92 / y

Solution: We have to solve for y, therefore, we will rearrange the given equation for y.
We get,
y = 92 ÷ cos5°
Using the calculator,
Cos5° = 0.99
Now substitute the value of cos5° into the main equation.
we get,
y = 92 ÷ 0.99
y = 92.92 (rounded to the nearest hundredth)

Q5: Given the shape attached, therefore, using the triangle given, we have:
Angle of elevation = 35°
length of Opposite side to the angle = x
Length of Hypoteneus = 12
Calculations:
Using the SOH CAH TOA rules:
SOH stands for SineФ = Opposite ÷ Hypotenuse.

CAH stands for CosineФ = Adjacent ÷ Hypotenuse.

TOA stands for TangentФ = Opposite ÷ Adjacent.

Hence,

SineФ = Opposite ÷ Hypotenuse

Substituting the values:

Sine35° = x ÷ 12

0.5735 = x ÷ 12

x = 0.5735 × 12

x = 6.88 (rounded to the nearest hundredth)

Q6: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 54°
length of the adjacent side to the angle = x
Length of Hypoteneus = 30
Calculations:
Using the SOH CAH TOA rules:

Hence,

CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

Cos54° = x ÷ 30

0.5877 = x ÷ 30

x = 0.5877 × 30

x = 17.63 (rounded to the nearest hundredth)

Q7: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 22°
length of the adjacent side to the angle = 85
length of the opposite side to the angle = x
Calculations:

Using the SOH CAH TOA rules:

Hence,

TangentФ = Opposite ÷ Adjacent

Substituting the values:

tan22° = x ÷ 85

0.4040 = x ÷ 85

x = 0.4040 × 85

x = 34.34 (rounded to the nearest hundredth)

Q8: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 16°
length of the opposite side to the angle = x
Length of Hypoteneus = 14
Calculations:
Using the SOH CAH TOA rules:
Hence,

CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

Sine16° = x ÷ 14

0.2756 = x ÷ 14

x = 0.2756 × 14

x = 3.86 (rounded to the nearest hundredth)

Q9: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 65°
length of the adjacent side to the angle = 9
length of the opposite side to the angle = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

TangentФ = Opposite ÷ Adjacent

Substituting the values:

tan65° = x ÷ 9

2.1445 = x ÷ 9

x = 2.1445 × 9

x = 19.30 (rounded to the nearest hundredth)

Q10: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 51°
length of the adjacent side to the angle = x
Length of Hypoteneus = 70
Calculations:
Using the SOH CAH TOA rules:
Hence,

CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

Cos51° = x ÷ 70

0.6293 = x ÷ 70

x = 0.6293 × 70

x = 44.05 (rounded to the nearest hundredth)

Q11: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 36°
length of the opposite side to the angle = 15
Length of Hypoteneus = x
Calculations:
Using the SOH CAH TOA rules:
Hence,

CosineФ = Adjacent ÷ Hypotenuse

Substituting the values:

Sine36° = 15 ÷ x

0.5877 = 15 ÷ x

x = 15 ÷ 0.5877

x = 25.52 (rounded to the nearest hundredth)

Q12: Given the shape attached, therefore, using the triangle given, we have:

Angle of elevation = 60°
length of the adjacent side to the angle = x
length of the opposite side to the angle = 100

Calculations:
Using the SOH CAH TOA rules:

Hence,

TangentФ = Opposite ÷ Adjacent

Substituting the values:

tan65° = 100 ÷ x

2.1445 = 100 ÷ x

x = 100 ÷ 2.1445

x = 46.63 (rounded to the nearest hundredth)

Q13: When a 25-ft ladder is leaned against a wall, it makes a 72° with the ground. How high up on wall does the ladder reach?

Solution: Given the shape attached, therefore, using the triangle given, we have:

The angle of elevation from the ground = 72°
length of the wall opposite to the angle = X
Length of ladder (Hypoteneus) = 25 feet

Calculations:
Using the SOH CAH TOA rules:
Hence,

SineФ = Opposite ÷ Hypotenuse

Substituting the values:

Sine72° = x ÷ 25

0.9510 = x ÷ 25

x = 25 ÷ 0.9510

x = 23.77 (rounded to the nearest hundredth)

ANSWERS TO QUESTION 14 AND 15 ARE ATTACHED

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3 years ago