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pantera1 [17]
2 years ago
5

A ladder is propped against the side of a building making a 30 degree

Mathematics
1 answer:
Dahasolnce [82]2 years ago
3 0
Using trigonometry. This picture shows a ‘not to scale’ sketch of the following question. Use Sin 30. Sin 30= opposite/hypo tenuse. That is simply dividing 6 ft over Sin 30 that is 12 feet.

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Solve for the variable in 6/18=x/36
liberstina [14]

6/18=x/36

18x=216

x=12

6 0
3 years ago
19) In the xy-plane, what is the y-intercept of the graph of the equation<br> y=3(x-1)(x+2)?
Dahasolnce [82]

Answer:

- 6

Step-by-step explanation:

Given

y = 3(x - 1)(x + 2) ← expand factors using FOIL

  = 3(x² + x - 2) ← distribute by 3

  = 3x² + 3x - 6

To find the y- intercept let x = 0, thus

y = 3(0)² + 3(0) - 6 = 0 + 0 - 6 = - 6

Thus y- intercept = - 6 ⇒ (0, - 6 )

3 0
3 years ago
Rex, Paulo, and Ben are standing on shore watching for dolphins. Paulo sees one surface directly in front of him about a hundred
Ksivusya [100]

1. m\angle BAC=m\angle CAD,\ m\angle ACB=m\angle ADC=90^{\circ}, then m\angle ABC=m\angle ACD and triangles ADC and ACB are similar by AAA theorem.


2. The ratio of the corresponding sides of similar triangles is constant, so


\dfrac{AC}{AB}= \dfrac{AD}{AC}.


3. Knowing lengths you could state that \dfrac{b}{c}= \dfrac{e}{b}.


4. This ratio is equivalent to b^2=ce.


5. m\angle ABC=m\angle CBD,\ m\angle ACB=m\angle CDB=90^{\circ}, then m\angle BAC=m\angle BCD and triangles BDC and BCA are similar by AAA theorem.


6. The ratio of the corresponding sides of similar triangles is constant, so


\dfrac{BC}{BD}= \dfrac{AB}{BC}.


7. Knowing lengths you could state that \dfrac{a}{d}= \dfrac{c}{a}.


8. This ratio is equivalent to a^2=cd.


9. Now add results of parts 4 and 8:


b^2+a^2=ce+cd.


10. c is common factor, then:


b^2+a^2=c(e+d).


11. Since e+d=c you have a^2+b^2=c\cdot c=c^2.



7 0
3 years ago
Read 2 more answers
16 is 40% of what number?
saw5 [17]
Sixteen is 40% of 40
7 0
3 years ago
Read 2 more answers
HELP ASAP!!! PLEASE CHECK ALL OF THEM!
pochemuha

Answer:

2.

A. (P+h)(x)

2x/x+4 (x-1) + x/x-1 (x+4)

2x^2-1/x^2-4

+

X^2+4/x^2-4

= 3x^2+3/x^2-4

B. (F-g)(x)

X^2-7x+6-x - 6

= x^2 -8x

C. (Fg)(x)

(X^2-7x+6)(x-6)

= x^3-13x^2+48x-36

D. (H/p)(x)

X/x-1 / 2x/x+4

X/x-1 / x+4/2x

= X^2+4x/2x^2-2x

3.

A. (F+g)(3)

X^2+1 + x-4

3^2+1 + 3-4

10 -1

= 9

B. (f-g)(0)

X^2+1 - x-4

0+1 -0-4

1-4

= -3

C. (Fg)(-k)

(X^2+1) (x-4)

(-k^2+1) (-k-4)

K^3+4k^2-k-4

D. (F/g)(k-2)

X^2+1 /x-4

K-2^2+1 / k-2 -2

= K^2-4k+5 / k-4

Step-by-step explanation:

8 0
3 years ago
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