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2 years ago

H S 45° 299 N 106 E 29° А A A

Mathematics

1 answer:

tankabanditka [31]2 years ago

3 0

Answer:

Can you pls write it straight

Step-by-step explanation:

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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
$sin \theta = -0.69178$

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2 years ago

Which is f(6) for the quadratic function graphed? 0 -2 * -0.5 1.5 4

rusak2 [61]

Answer:

Step-by-step explanation:

I'm pretty sure theres a graph to this so....

The given parabolic graph of quadratic function

To find:What is F(6) for the quadratic function

Solution:We have given graph of quadratic function

there is a different value for Y and a different value for X

For f(6)

here X=6, need to find graph of quadratic function

X=6

f(6)=4

Y=4

Your welcome :)

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2 years ago

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Nuetrik [128]

Answer:

${ \tt{ \tan(x) = \frac{opposite}{adjacent} }} \\ \\ { \tt{ \tan( \theta) = \frac{30}{16} }}$

5 0

2 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

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3 years ago

A student solved this problem and said the answer is miles.

Gre4nikov [31]

B. No, the answer is not reasonable. It should be about 2 miles.

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2 years ago

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