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Bond [772]
2 years ago
10

1) A chemist has 250 ml of a 5% acid solution. How much water should be added to this solution to reduce it to a 3% solution.

Mathematics
1 answer:
Firlakuza [10]2 years ago
3 0

Answer:

153.1ml

Step-by-step explanation:

We have 12.5ml of acid in the solution.

If we want this to be 3% then we can divide this value by 3 and times by 100 to get the total value.

416.6ml total solution amount

415.6-12.5-250=153.1

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A parallelogram has sides 24m and 5m. The height corresponding to a 24m base is 4m. What is the height corresponding to a 5m bas
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An isosceles triangle ABC with the base BC is inscribed in a circle. Find the measure of angles of it, if measure of arc BC = 10
Usimov [2.4K]

Answer:

The measures of the angles of the triangle are 51° , 64.5° , 64.5°

OR

The measures of the angles of the triangle are 129° , 25.5° , 25.5°

Step-by-step explanation:

* Lets explain the meaning of the inscribed triangle in a circle

- If a triangle inscribed in a circle, then the vertices of the triangle lie

 on the circumference of the circle and each vertex is an inscribed

 angle in the circle subtended by the opposite arc

- Fact in the circle the measure of the inscribed angle is 1/2 the

 measure of its subtended arc

* Now lets solve the problem

- Δ ABC is an isosceles with the base BC

∴ AB = AC

∴ m∠B = m∠C

- Δ ABC is inscribed in a circle

∴ ∠A is inscribed angle subtended by arc BC (minor or major)

# The measure of the minor arc is less than 180° and the measure of

  the major arc is greater then 180° and the sum of the two arcs

  equals the measure of the circle which is 360°

∴ ∠B subtended by arc AC

∴ ∠C subtended by arc AB

∵ The measure of the arc BC = 102°

- There is two cases in this question

(1) If the angle A subtended by the minor arc BC

(2) If the angle A subtended by the major arc BC

- Lets solve case (1)

∵ ∠A is an inscribed angle subtended by the minor arc BC

∴ m∠A = 1/2 the measure of the arc BC

∵ The measure of the arc BC is 102°

∴ m∠A = 1/2 × 102 = 51°

∵ The sum of the measures of the interior angles of a triangle is 180°

∴ m∠A + m∠B + m∠C = 180°

∴ 51 + m∠B + m∠C = 180° ⇒ subtract 51 from both sides

∴ m∠B + m∠C = 129°

∵ m∠B = m∠C ⇒ isosceles Δ

∴ m∠B = m∠C = 129/2 = 64.5°

* The measures of the angles of the triangle are 51° , 64.5° , 64.5°

- Lets solve case (2)

∵ ∠A is an inscribed angle subtended by the major arc BC

∴ m∠A = 1/2 the measure of the arc BC

∵ The measure of the minor arc BC is 102°

∵ The measure of the circle is 360°

∴ The measure of the major arc = 360 - 102 = 258°

∴ m∠A = 1/2 × 258 = 129°

∵ The sum of the measures of the interior angles of a triangle is 180°

∴ m∠A + m∠B + m∠C = 180°

∴ 129 + m∠B + m∠C = 180° ⇒ subtract 129 from both sides

∴ m∠B + m∠C = 51°

∵ m∠B = m∠C ⇒ isosceles Δ

∴ m∠B = m∠C = 51/2 = 25.5°

* The measures of the angles of the triangle are 129° , 25.5° , 25.5°

4 0
2 years ago
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