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3 years ago

7

Geometry help please! what is the height of the sign?please and thank u!

1 answer:

valentina_108 [34]3 years ago

6 0

8ft. I hope this helped. .-.

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The question is in the file.

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3 years ago

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Find the absolute value.<br><br> 1 -31=___<br><br> -1 121=___<br><br> -1 -221=___

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3 years ago

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Need help on these trig homework

allochka39001 [22]

8. The length of the support is 7. 9 m

9. The length of the conveyer is 12m

3. a = 59m

A = 44°

B = 52°

4. c = 88. 6 mm

b = 49. 1 m

B = 34°

<h3>How to solve the trigonometry</h3>

8. We have the angle to be 20 degrees

Opposite side = x

hypotenuse = 23m

Using the sine ratio

sin θ = opposite/ hypotenuse

$sin 20 = \frac{x}{23}$

Cross multiply

$sin 20$ × $23 = x$

$x = 0. 3420$ × $23$

x = 7. 9 m

The length of the support is 7. 9 m

9. The angle of elevation is 37. 3 degrees

Hypotenuse = 19 . 0m

Opposite = x

Using the sine ratio

sin θ = opposite/ hypotenuse

$sin 37. 3 = \frac{x}{19}$

cross multiply

$x = 0. 6059$ × $19$

x = 11.5

x = 12 m in 2 significant figures

The length of the conveyer is 12m

3. To determine the sides and angles, we use the sine rule;

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

For side a, we use the Pythagorean theorem

$c^2 = a^2 + b^2$

$85^2 = a^2 + 67^2$

$a = \sqrt{89^2-67^2}$

$a = \sqrt{3432}$

a = 58. 58, a = 59m

To find angle A and B, use the sine rule

$\frac{59}{sin A } = \frac{85}{sin 90}$

cross multiply

$sin A$ × $85$ = $sin 90$ × $59$

make sin A subject of formula

$sin A = \frac{59}{85}$

$sin A = 0. 6941$

A = $sin^-^1(0. 6941)$

A = 44°

$\frac{67}{sin B} = \frac{85}{sin 90}$

cross multiply

$sin B$ × $85$ = $sin 90$ × $67$

make sin b subject of formula

$sin B = \frac{67}{85}$

$sin B = 0. 7882$

B = $sin^-^1( 0. 7882)$

B = 52°

4. To find the sides, we use the sine rule;

$\frac{74. 0}{sin 56. 6} = \frac{c}{sin 90}$

Cross multiply

$sin 56. 6$ × $c = sin 90$ × $74$

make 'c' subject of formula

$c = \frac{74}{0. 8348}$

c = 88. 6 mm

To find length b, we use the Pythagorean theorem

$c^2 = a^2 + b^2$

$b^2 = c^2 - a^2$

$b^2 = 88. 8^2 - 74^2$

$b = \sqrt{7885. 44 - 5476}\\\\ b = \sqrt{2409. 44}$

b = 49. 1 m

$\frac{74. 0}{sin 56. 6} = \frac{49. 1}{sin B}$

cross multiply

$sin B = \frac{40. 99}{74. 0}$

B = $sin^-^1(0. 5539)$

B = 34°

Learn more about trigonometric identity here:

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3 0

2 years ago

How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and con

sladkih [1.3K]

65 sequences.

Lets solve the problem,

The last term is 0.

To form the first 18 terms, we must combine the following two sequences:

0-1 and 0-1-1

Any combination of these two sequences will yield a valid case in which no two 0's and no three 1's are adjacent

So we will combine identical 2-term sequences with identical 3-term sequences to yield a total of 18 terms, we get:

2x + 3y = 18

Case 1: x=9 and y=0

Number of ways to arrange 9 identical 2-term sequences = 1

Case 2: x=6 and y=2

Number of ways to arrange 6 identical 2-term sequences and 2 identical 3-term sequences =8!6!2!=28=8!6!2!=28

Case 3: x=3 and y=4

Number of ways to arrange 3 identical 2-term sequences and 4 identical 3-term sequences =7!3!4!=35=7!3!4!=35

Case 4: x=0 and y=6

Number of ways to arrange 6 identical 3-term sequences = 1

Total ways = Case 1 + Case 2 + Case 3 + Case 4 = 1 + 28 + 35 + 1 = 65

Hence the number of sequences are 65.

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8 0

2 years ago

The school's basketball team lost 4 times as many games as they won. They also tied 5 fewer games than they won.

jekas [21]

Let w won games, t tied games, s lose games

s=4w

t=w-5

s+t+w=43

4w+(w-5)+w=43

6w=48

w=7

s=4*7=21 games lost

7 0

3 years ago