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3 years ago

9

In circle R, RD(line over it) is a radius and DF (line over it) is a tangent segment which statement must be true?

1 answer:

butalik [34]3 years ago

5 0

We'll have an attached picture (Ignore the labeling). We can easily observe that DF (in the picture BC), cannot be equal to the radius. Then, we cannot also claim that RD=RF. And since we are talking about the tangent segment, we'll have a right triangle. The correct answer is C)

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Convert 2845 inches per second to miles per<br> hour. Round to one decimal place.

garik1379 [7]

the answer would be 161.6 mph

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3 years ago

A ski rental store rents a pair of skis for $44 a day and a snowboard for $58 a day. Yesterday, the store made $2,232 on ski and

Katen [24]

First, let the number of skis rented by x and the number of snowboards rented by y. We can then assemble the first equation from the amount of money made from the rentals.

44x + 58y = 2232

The second equation can come from the fact that 9 more skis were rented than snowboards.

y = x - 9

Therefore our system is:

44x + 58y = 2232

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4 years ago

How much material will Dmitri's mom need for the tent, including the floor?

suter [353]

<h3>Answer: 21.4 square meters</h3>

Work Shown:

A = area of the front triangular face

A = base*height/2

A = 2*1.7/2

A = 1.7

L = lateral area

L = (perimeter of the base)*(depth)

L = (2+2+2)*(3)

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SA = total surface area of a triangular prism

SA = 2*A+L

SA = 2*1.7+18

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7 0

2 years ago

Read 2 more answers

Prove that a triangle with the sides a - 1 cm to root under a

vivado [14]

Question is Incomplete, Complete question is given below.

Prove that a triangle with the sides (a − 1) cm, 2√a cm and (a + 1) cm is a right angled triangle.

Answer:

∆ABC is right angled triangle with right angle at B.

Step-by-step explanation:

Given : Triangle having sides (a - 1) cm, 2√a and (a + 1) cm.

We need to prove that triangle is the right angled triangle.

Let the triangle be denoted by Δ ABC with side as;

AB = (a - 1) cm

BC = (2√ a) cm

CA = (a + 1) cm

Hence,

${AB}^2 = (a -1)^2$

Now We know that

$(a- b)^2 = a^2+b^2 - 2ab$

So;

${AB}^2= a^2 + 1^2 -2\times a \times1$

${AB}^2 = a^2 + 1 -2a$

Now;

${BC}^2 = (2\sqrt{a})^2= 4a$

Also;

${CA}^2 = (a + 1)^2$

Now We know that

$(a+ b)^2 = a^2+b^2 + 2ab$

${CA}^2= a^2 + 1^2 +2\times a \times1$

${CA}^2 = a^2 + 1 +2a$

${CA}^2 = AB^2 + BC^2$

[By Pythagoras theorem]

$a^2 + 1 +2a = a^2 + 1 - 2a + 4a\\\\a^2 + 1 +2a= a^2 + 1 +2a$

Hence, ${CA}^2 = AB^2 + BC^2$

Now In right angled triangle the sum of square of two sides of triangle is equal to square of the third side.

This proves that ∆ABC is right angled triangle with right angle at B.

4 0

3 years ago

1997×99 in vertical order

Oksi-84 [34.3K]

Answer:

197703

Step-by-step explanation:

1

9

7

7

0

3

8 0

3 years ago