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3 years ago

Can someone help me?

Mathematics

1 answer:

hoa [83]3 years ago

5 0

Answer:

Step-by-step explanation:

If arccos (√3 / 2) = β, then cos β = adjacent side / hypotenuse = √3 / 2. This tells us immediately that the adjacent side length is √3 and the hypotenuse is 2. Via the Pythagorean Theorem, we know that the third side is 1.

This information rules out the 1st and 3rd answer choices.

Because cos 30° = adj / hyp = √3 / 2, we can conclude that the correct answer is the 4th one: 30°, 60°, 90°

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Parallel to y=-1/4x-12 through (12,10)

Yanka [14]

Y=-1/4x+13 not sure though

8 0

3 years ago

Which expression is equivalent to n2 + 26n + 88 for all values of n?

Lelechka [254]

Answer:

Option B is correct.

The expression which is equivalent to $n^2+26n+88$ is; $(n+22)(n+4)$

Explanation:

Given the equation: $n^2+26n+88$ for all values of n.

The quadratic equation in this form; $ax^2+bx+c =0$

First find two numbers that multiply to give ac and add to give b.

From the given equation;

a =1 , b = 26 and c =88

then,

the two number that multiply to give ac =88 is, 22 or 4

and they add up to give b=26 (i.e 22+4)

Now, rewrite the middle term i.e 26n with 22n and 4n , we have

$n^2+22n+4n+88=0$

Now, factor the first two terms and last two terms,

$n(n+22)+4(n+22)$

we see that $(n+22)$ is common to both terms so, we have;

$(n+22)(n+4)$

Therefore, the expression $(n+22)(n+4)$ is equivalent to $n^2+26n+88$

Check:

(n+4)(n+22) = $n\cdot n+ 22n+4n+88$ = $n^2+26n+88$ [ True]

5 0

3 years ago

Read 2 more answers

Solve for y a(n+y)=10y+32 what is the answer

Solnce55 [7]

A (n + y) = 10y + 32
(an + ay) = 10y + 32

an + ay = 32 + 10y

Solve for "a"
-32 + an + ay + (-10y) = 32 + 10y + (-32) + (-10y)
-32 + an + ay + -10y = 32 + -32 + 10y + -10y
<span>- 32 + an + ay + (-10y) = 0 + 10y + (-10y)
- 32 + an + ay + (-10y) = 10y + (-10y)

</span><span>10y + -10y = 0
-32 + an + ay + (-10y) = 0

Thi could not be determined. (no solution)</span>

6 0

3 years ago

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t

WINSTONCH [101]

Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

$\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}$

Seperate the differential equation and solve for the constant C.

$\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2} }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}$