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coldgirl [10]
3 years ago
9

What is the solution -1 -7 ?

Mathematics
1 answer:
Effectus [21]3 years ago
3 0

Answer:

-8

Step-by-step explanation:

Add the two numbers but maintain the negative sign.

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Given that
Iteru [2.4K]

Answer:

y=3÷9

Step-by-step explanation:

2x-9y=11

or,2×7-9y=11

or,14-11=9y

or,3÷9=y

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3 years ago
Second​ dose, 122 of 676 subjects in the experimental group​ (group 1) experienced fever as a side effect. After the second​ dos
Oksanka [162]

Answer:

Step-by-step explanation:

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3 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
Expand 3(t -2) 3(t-2)=
iogann1982 [59]

Answer:

I hope this helps:)

Step-by-step explanation:

3 0
3 years ago
PLEASE HELP MATH 10+ POINTS PLEASE
tankabanditka [31]

Answer:

749

Step-by-step explanation:

8349

3 0
3 years ago
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