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3 years ago

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A circular track will be built around a square playing field that is 100 ft x 100 ft. What is the area of the smallest track tha

t can fit outside of the
playing field?
2501 ft2
5,000 TI in
1,000
in 2
2,500 T in2

1 answer:

Anna35 [415]3 years ago

4 0

Answer:

2501 ft2

Step-by-step explanation:

The other answers are in in2 or in. In this case you would not be looking for an answer in in2 or in.

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Solve the right triangle, ΔABC, for the missing sides and angle to the nearest tenth given angle B = 39° and side c = 13.

hram777 [196]

Answer:

Part 1) $b=8.2\ units$

Part 2) $a=10.1\ units$

Part 3) $A=51\°$ and $C=90\°$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

Find the side b

we know that

In the right triangle ABC

The function sine of angle B is equal to divide the opposite side angle B (AC) by the hypotenuse (AB)

$sin(B)=AC/AB$

we have

$AB=c=13\ units$

$AC=b$

$B=39\°$

substitute

$sin(39\°)=b/13$

solve for b

$b=(13)sin(39\°)$

$b=8.2\ units$

step 2

Find the side a

we know that

In the right triangle ABC

The function cosine of angle B is equal to divide the adjacent side angle B (BC) by the hypotenuse (AB)

$cos(B)=BC/AB$

we have

$AB=c=13\ units$

$BC=a$

$B=39\°$

substitute

$cos(39\°)=a/13$

solve for a

$a=(13)cos(39\°)$

$a=10.1\ units$

step 3

Find the measure of angle A

we know that

In the right triangle ABC

$C=90\°$ ----> is a right angle

$B=39\°$

∠A+∠B=90° ------> by complementary angles

substitute the given value

$A+39\°=90\°$

$A=90\°-39\°$

$A=51\°$

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4 years ago

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Answer:

B, D, and E are all equal to the equation above.

Step-by-step explanation:

all of them equal 5 and 4/5

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3 years ago

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I hope this helps you

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3 years ago

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Find the least common multiple<br> (LCM) of 7, 9, and 21.

prisoha [69]

Answer: 63

Explanation: trial and error of comparing

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2 years ago

I’ll give brainliist

Luda [366]

Answer:

JK = 6

NL = 4

Step-by-step explanation:

J > K > M > N > L

JN = 14

KM= 4

MN = 4

LN = 4

JN - KM -MN= JK

14-4-4=JK

JK = 6

if 3 measures are 4 then this is showing radius split into two sectors

JK therefore is segment 1 = 6

KM therefore is segment 2 = 4

MN therefore is the sector radius from M = 4

NL therefore is the sector radius from N to L which meets J and = 4 as it is an end line of the segment.

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3 years ago