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3 years ago

Add the two expressions. 3z−43z−4 and 2z + 5

Mathematics

1 answer:

yuradex [85]3 years ago

3 0

Solution:

1) Gather like terms
(3z-43z+2z)

2) Simplify
-38z-4+5

3) Simplify
-38z+1

Done!

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Solve the triangle. A = 54°, b = 11, c = 8

Delicious77 [7]

In this problem, you apply principles in trigonometry. Since it is not mentioned, you will not assume that the triangle is a special triangle such as the right triangle. Hence, you cannot use Pythagorean formulas. The only equations you can use is the Law of Sines and Law of Cosines.

For finding side a, you can answer this easily by the Law of Cosines. The equation is

a2=b2 +c2 -2bccosA
a2 = 11^2 + 8^2 -2(11)(8)(cos54)
a2 = 81.55
a = √81.55
a = 9

Then, we use the Law of Sines to find angles B and C. The formula would be

a/sinA = b/sinB = c/sinC
9/sin54° = 11/sinB
B = 80.4°

9/sin54° = 8/sinC
C = 45.6°

The answer would be: a ≈ 9, C ≈ 45.6, B ≈ 80.4

5 0

3 years ago

How many tenths are in 8.32?

Finger [1]

there are 3 tenths in 8.32

8 0

2 years ago

Read 2 more answers

World Toy buys bicycles for $40 and sells them for $95. What is the percent mark-up in the price?

Arisa [49]

Answer:

57%

Step-by-step explanation:

4 0

3 years ago

Given |u| = 2.5, |v| = 3.2, and the angle between the vectors is 60°, find the value of u · v?

Alenkasestr [34]

Step-by-step explanation:

u.v=|u||v|cos60°

u.v=(2.5)(3.2)(1/2)

u.v=(8)(1/2)

u.v=4

4 0

3 years ago

A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a

Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 = 0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = $2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750$

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

$W_2 = \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5$

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0

2 years ago