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3 years ago

Write the ratios for sin A and cos B. The triangle is not drawn to scale.

Mathematics

1 answer:

babymother [125]3 years ago

4 0

By definition we have:
sineA = (C.O) / (h)
cosA = (C.A) / (h)
Where,
C.O: opposite leg
C.A: adjacent leg
h: hypotenuse
Substituting values:
sinA = (48) / (50)
cosA = (14) / (50)
Answer:
sineA = (48) / (50)
cosA = (14) / (50)
Option 2

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Are 5/3 and 1 2/3 equivalent?

Semmy [17]

Answer:

Yes.

Step-by-step explanation:

3/3 is equal to 1. If you add 3/3 to 2/3, you get 5/3, which means that yes, they are equivalent.

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2 years ago

Read 2 more answers

Solve -7<3x+2<5 (Multiple choice, picture added)

pishuonlain [190]

$\\ \sf\longmapsto -7$

Substract 2 from all sides

$\\ \sf\longmapsto -7-2$

$\\ \sf\longmapsto -9$

Divide all sides by 3

$\\ \sf\longmapsto \dfrac{-9}{3}$

$\\ \sf\longmapsto -3$

Option A is correct

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2 years ago

Read 2 more answers

Write the first five terms of the arithmetic sequence. First term: -10 and common difference: 2

Sholpan [36]

Answer:

The first five terms of the arithmetic sequence.

-10 , -8 , -6 , -4 ,-2 ..........

Step-by-step explanation:

Explanation

Given that the first term is -10 and common difference d = 2

we know that the sequence in A.P

a ,a+d ,a+2d,a+3d ,a+4d........... are in AP

-10, -10+2, -10+2(2), -10+3(2) , -10+4(2)..................

-10 , -8 , -6 , -4 ,-2 ..........

Final answer:-

The first five terms of the arithmetic sequence.

-10 , -8 , -6 , -4 ,-2 ..........

6 0

2 years ago

Please answer CORRECTLY !!!!! Will mark brainliest !!!!!!!!!!

VashaNatasha [74]

Answer:

5x² + 10

Step-by-step explanation:

5(x² + 2)

= 5x² + 10

6 0

2 years ago

The fish population in a pond with carrying capacity 1200 is modeled by the following logistic equation where N(t) denotes the n

MArishka [77]

9514 1404 393

Answer:

(dN)/(dt) = (0.4)/(1200)N(1200-N) -50

142 fish

Step-by-step explanation:

A) The differential equation is modified by adding a -50 fish per year constant term:

(dN)/(dt) = (0.4)/(1200)N(1200-N) -50

B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.

(0.4/1200)(N)(1200-N) -50 = 0

N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient

N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square

(N -600)^2 = 210,000 . . . . . simplify

N = 600 + √210,000 ≈ 1058

This steady-state number of fish is ...

1200 - 1058 = 142 . . . . below the original carrying capacity

8 0

2 years ago