2. Simplify each expression.<br> A.5+5 (24 +12)3

Please help me with number ten and twelve

jasenka [17]

Answer: $93,\ 34^{\circ}$

Step-by-step explanation:

Given

For figure (i) both triangles are congruent

$\therefore 8x-27=4x+33\\\Rightarrow 8x-4x=33+27\\\Rightarrow 4x=60\\\Rightarrow x=15$

$\Rightarrow PQ=8x-27\\\Rightarrow PQ=93$

For figure (ii)

Two triangles WXZ and WZY are congruent

$\therefore 9x-28=5x-8\\\Rightarrow 4x=20\\\Rightarrow x=5^{\circ}\\\Rightarrow \angle YWX=2(9x-28)\\\Rightarrow \angle YWX=2(17)\\\Rightarrow \angle YWX=34^{\circ}$

5 0

2 years ago

The measure of the supplement of an angle is seven times as large as the measure of its complement find the measure of the angle

8090 [49]

Answer: original angle = 75°

Its supplement 105°

Its complement = 15°

Step-by-step explanation:

We know that,

Sum of supplementary angles is 180 degrees.
Sum of complementary angles is 90 degrees.

Let x be the original angle , then its supplement = 180° - x

its complement = 90° - x

As per given,

180° - x = 7(90° - x)

⇒ 180° - x = 630° - 7x

⇒ 7x- x = 630° -180°

⇒ 6 x = 450°

⇒ x = 75° [divide both sides by 6]

So, original angle = 75°

Its supplement = 180° - 75° = 105°

Its complement = 90° -75° = 15°

Hence, original angle = 75°

Its supplement 105°

Its complement = 15°

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

Please help me with these two questions. Find the circumference AND area for the circles on the right. (I will mark brainliest)

notsponge [240]

5). Radius = 6 km. Circumference = 12π km. Area = 36π km²

6). Diameter = 8 mm. Circumference = 8π mm. Area = 16π mm²

7 0

2 years ago

Read 2 more answers

a report in a newspaper or magazine, or on the Internet that uses results from a survey. Evaluate how the survey uses ratios to

gavmur [86]

Answer:

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Step-by-step explanation:

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5 0

2 years ago

2. Simplify each expression. A.5+5 (24 +12)3