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2 years ago

WHO NEED POINTS< COME AND GET POINTS!!!!!

Mathematics

2 answers:

KiRa [710]2 years ago

6 0

Answer:

thank you by a ton

Step-by-step explanation:

Vaselesa [24]2 years ago

3 0

Answer:

Wow Tysm!

Step-by-step explanation:

TYSM

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Verify this trig function. <br> tanysiny = secy-cosy ?

avanturin [10]

$LHS\\ \\ =\tan { y } \sin { y } \\ \\ =\frac { \sin { y } }{ \cos { y } } \cdot \sin { y }$

$\\ \\ =\frac { \sin ^{ 2 }{ y } }{ \cos { y } } \\ \\ =\frac { 1-\cos ^{ 2 }{ y } }{ \cos { y } }$

$\\ \\ =\frac { 1 }{ \cos { y } } -\frac { \cos ^{ 2 }{ y } }{ \cos { y } } \\ \\ =\sec { y } -\cos { y } \\ \\ =RHS$

6 0

3 years ago

HELP PLZ PICTURE^ Need it asap porfa

timurjin [86]

Answer:

it has one solution

Step-by-step explanation:

1.y=x-3

2. 3y-3x sub x-3 in place of y therefore

it can also be written as 3x-3x-9=9

if you add 9 to both sides 3x-3x-9+9=-9+9

0+0=0

3 0

3 years ago

You want to wrap a gift shaped like the regular triangular prism shown. How many square inches of wrapping paper do you need to

Tju [1.3M]

This would divide the x to 7 to get 86

8 0

2 years ago

HELP FAST 100 POINTS 100 POINTS What is the value of the expression when a = 5, b = 4 , and c = 6?⁢ 4a2b−c 2 5 9 10

ohaa [14]

For this case we must find the value of the following expression:

$4a ^ 2b-c$

When:

$a = 5\\b = 4\\c = 6$

Substituting the given values we have:

$4 (5) ^ 2 * 4-6 =\\4 (25) (4) -6=\\400-6 =\\394$

Finally, the value of the expression is 394.

Answer:

The value of the expression is 394.

3 0

2 years ago

Read 2 more answers

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),

pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

$\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}$

$(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}$

When t=0, S(t)=15

$15=30+Ce^{-k*0}\\C=15-30\\C=-15$

When t = 2, S(t)=20

$20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}$

Therefore:

$S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22$

8 0

2 years ago