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1 year ago

HELPPPPP!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

monitta1 year ago

4 0

Answer: All the sides given are complete so we can use the information given to find 1 triangle!!

Olin [163]1 year ago

3 0

Answer:

Step-by-step explanation:

All the sides are complete, hence we can use the information given to find just 1 triangle

Triangles

From the given information, we have that:

m<B = pi/6 = 30 degres

c = 10

b = 5

This shows that we can use the cosine rule to find the third side b of the triangle. Since all the sides are complete, hence we can use the information given to find just 1 triangle

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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

Directions: Simplify the following expressions using the imaginary number.

drek231 [11]

Answer:

1 i(sqrt5)

2 7i

3 -9i

4 -64i

5 -5i(sqrt2)

6 -4i(sqrt3)

7 10i(sqrt3)

8 2i(sqrt2)

9 3i(sqrt7)

10 -30i(sqrt2)

Step-by-step explanation:

i is the square root of (-1). We can simplify each one as such.

5 0

1 year ago

PLEASE HELP ASAP!!! (No links)

marta [7]

Answer:

No it is not a solution

Step-by-step explanation:

To solve this, you should replace your (x,y) with (8,3)- 8 being your x value and 3 being your y value.

So this inequality, your y value is supposed to be bigger than the value of whatever 3x-4 equals.

So, your inequality should be 3>3(8)-4

3*8=24-4=20

So is 3>(greater than)20? No. Which then means that (8,3) is not a solution because it would make the inequality not true.

Hope this helps!

6 0

2 years ago

Person A can paint the neighbor's house 6 times as fast as Person B. The year A and B worked together it

uranmaximum [27]

Person A takes 2.3 days to paint the house working alone

Person B takes 13.8 days to paint the house working alone

<em><u>Solution:</u></em>

Let "x" be the number of days it takes person A to paint the house

Person A can paint the neighbor's house 6 times as fast as Person B

Therefore.

Number of days it takes person B to paint the house = 6x

Their rates of painting are added to get the rate working together

$\frac{\text{1 house}}{\text{x days}} + \frac{\text{1 house}}{\text{6x days}} = \frac{\text{1 house}}{\text{2 days}}$

$\frac{1}{x} + \frac{1}{6x} = \frac{1}{2}\\\\\frac{1 \times 6}{6x} + \frac{1}{6x} = \frac{1}{2}\\\\\frac{7}{6x} = \frac{1}{2}\\\\\ x = \frac{7}{3}= 2.3$

Thus person A takes 2.3 days to paint the house working alone

Person B = 2x = 6(2.3) = 13.8

Thus person B takes 13.8 days to paint the house working alone

3 0

3 years ago

If you’re good with permutations in math 30 please help and answer this easy question below real answers only

vredina [299]

Answer:

Hello,

c) 720

Step-by-step explanation:

consonants : 5

5 x 4 x 3 x 2 x 1 = 120

other letters : 3

3 x 2 x 1 = 6

and 6 x 120 = 720

720 arrangements here

3 0

3 years ago