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3 years ago

-4 = r/20 -5Solve for r. Show your steps

Mathematics

2 answers:

Delvig [45]3 years ago

3 0

-4 = r/20 - 5

Add 5 to both sides

1 = r/20

Multiple by 20 on both sides

20 = r

Paladinen [302]3 years ago

3 0

Answer:

r = 20

Explanation:

-4 = r/20 - 5

Simplify both sides of the equation

-4 = r/20 - 5

-4 = 1/20r + -5

-4 = 1/20r - 5

Flip the equation

1/20r - 5 = -4

Add 5 to both sides

1/20r - 5 + 5 = -4 + 5

1/20r = 1

Multiply both sides of the equation by 20

(20)1/20r = 1(20)

r = 20

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Rama09 [41]

Answer:

C. 18 cm

Step-by-step explanation:

The ratio of the sides of the triangle shown is 12 : 15 = 4 : 5. We know it is a right triangle, so we know the missing side length completes the ratio

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Half of XY is 9 cm, so the length of the entire chord is 18 cm.

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The chord is tangent to the inner circle, so makes a 90° angle with the radius to that tangent point. This tells you that the triangle shown is a right triangle. It also tells you that the short radius bisects the chord. The Pythagorean theorem can be used to find the length of the side not shown (half the chord length).

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7 0

3 years ago

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.