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3 years ago

What is 15% of ____ is 27

Mathematics

2 answers:

Vinvika [58]3 years ago

5 0

Answer:

180

Step-by-step explanation:

180 * 15% = 27

Hope this Helps!!!

serious [3.7K]3 years ago

3 0

Answer:

it's 180

15% of 180 is 27

hope it helps :)

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Brianna evaluated the expressions 42+4 (6-2). Which number shows the correct solution.

TEA [102]

Answer:

Step-by-step explanation:

42+4(6-2) =

42+4(4) =

42+16 = 58

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3 years ago

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What is the solution of y= 4x + 6 when y= 18?

11Alexandr11 [23.1K]

Answer:

18 = 4(3)+6

12 + 6

Step-by-step explanation:

the answer is 18 = 4(3)+6

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I need your help guys !!!!!!!!

Ber [7]

$\bold{\huge{\blue{\underline{ Solution}}}}$

$\bold{\underline{ We \: have \:given \: that, }}$

Distance between the house and tower is 35 m
The height of the tower is 60 m
The height of the house is 25 m

$\bold{\underline{ Now}}$

Height of the house is 25m

$\bold{\underline{ So, }}$

$\sf{ CE = 25m }$

Therefore,

$\sf{ AB = AD - CE }$

$\sf{ AB = 60 - 25 }$

$\sf{ AB = 35 m }$

Now, In Right angled ABC

$\bold{tan Φ =}{\bold{\frac{ AB}{BC}}}$

$\bold{tan Φ =}{\bold{\frac{ 35}{35}}}$

$\bold{ tan Φ = 1 }$

$\bold{ tan Φ = 45 °}$

$\sf{\red{Hence, \: The \:angle \:of \:depression\: is \:45°}}$

7 0

2 years ago

Add 3 5/12 + 2 5/9 . Simplify the answer and write as a mixed number.

Kipish [7]

5 35/36, solving steps attached below

3 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