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Leokris [45]
3 years ago
11

gus wrote the following riddle : i am a number between 30 and 60 my ones digit is three less than my ones digit i am a prime num

ber
Mathematics
2 answers:
Sever21 [200]3 years ago
8 0
The numbers can be 31, 37, 41, 47, 53, and 59.Hope this helps
Vsevolod [243]3 years ago
3 0
47 is the only logical answer 7-3=4

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Find sin θ if θ is in Quadrant III and tan θ = . 0.958
Leokris [45]
Use the following identities:
sec^2 = 1  + tan^2 \\  \\ sec = \frac{1}{cos} \\  \\ sin^2 = 1 - cos^2
Also because the angle is in quadrant 3, sin must be negative.
Therefore
sin = - \sqrt{1 - \frac{1}{1 + tan^2}}
Subbing in tan = 0.958
sin \theta = -0.69178
7 0
3 years ago
The points plotted below are on the graph of a polynomial. Some of the roots to this polynomial are integers. Which of the follo
Vikki [24]
The only root is -2.  The roots are where the function goes through the x-axis.  This polynomial goes through in lots of places, but the only answer that matches is -2
8 0
3 years ago
A bucket holds b liters of water. But a jar holds 8 fewer liters than the bucket. How many times can the bucket hold than the ja
kykrilka [37]

Answer:

b / (b - 8)

Step-by-step explanation:

Given that :

Litres of water held by bucket = b

Jar holds 8 litres fewer than bucket; hence, litres held by jar = b - 8

The number of times bucket can hold than jar is :

Litres held by bucket / litres held by jar

b / (b - 8)

8 0
3 years ago
Fond the equation of the line shown
zloy xaker [14]

Answer:

y+-1/2x-1

Step-by-step explanation:

The line is negative

It rises one digit and runs 2

and it crosses though -1 on the y axis

4 0
2 years ago
Write the trigonometric expression sin(sin−1u−tan−1v) as an algebraic expression in u and v. Assume that the variables u and v r
igomit [66]

Answer:

[u – v√(1 – u²)]/√(1 + v²)

Step-by-step explanation:

Let sin^-1(u) = A, therefore sinA = u.

We know that sin(theta) = opposite/hypothenuse

Therefore, sinA = u/1 and u is the opposite side to angle A while 1 is the hypotenuse. Draw an acute triangle placing u opposite to angle A and 1 as the hypotenuse. By Pythagoras theorem the adjacent would be √(1 – u²).

By doing this, it means cosA = adjacent/hypotenuse = √(1 – u²)/1 = √(1 – u²)

Also, let tan^-1(v) = B, therefore tanB = v.

We know that tan(theta) = opposite/adjacent

Therefore, tanB = v/1 and v is the opposite side to angle B while 1 is the adjacent. Draw an acute triangle placing v opposite to angle B and 1 as the adjacent. By Pythagoras theorem the hypothenuse would be √(1 + v²).

Therefore, sinB = opposite/hypotenuse = v/√(1 + v²) and cosB = adjacent/hypotenuse = 1/√(1 + v²)

Now,

sin[sin^–1(u) – tan^–1(v)] =

sin(A – B) =

sinAcosB – sinBcosA =

u[1/√(1 + v²)] – [v/√(1 + v²)][√(1 – u²)] =

[u/√(1 + v²)] – [v√(1 – u²)/√1 + v²)] =

[u – v√(1 – u²)]/√(1 + v²).

8 0
3 years ago
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