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Reptile [31]
3 years ago
15

Find the terminal point p(x y) on the unit circle determined by the given value of t. t = -3 pi / 2

Mathematics
1 answer:
Vlada [557]3 years ago
7 0
-3pi/2 is the same thing as pi/2

the terminal point on the unit circle for pi/2 is (0,1)

so that means the terminal point for -3pi/2 is also (0,1).


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What is 52/624 simplified?<br><br>(Hurry please) ​
ZanzabumX [31]

Hey there!

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Simplifying fractions means dividing the numerator and the denominator by the same number.

So we know that both 52 and 624 are divisible by 2. However, it will take some time.

Turns out 624 is divisible by 52!

\frac{1}{12} is the answer. Hope it helps!

#HaveAnAmazingDay

GraceRosalia

8 0
2 years ago
Read 2 more answers
WILL GIVE BRAINLIEST
levacccp [35]

Answer:

a.  5^{2} x^{3}

b. 8^{3}

c. 4^{3} x^{4}

Step-by-step explanation:

a.  5^{2} x^{3}

b. 8^{3}

c. 4^{3} x^{4}

5 0
2 years ago
P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9
Feliz [49]

Answer:

x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > \frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1. Go ahead and simplify.

x=\frac{9}{d-1};\quad \:d\ne \:1.

Now, \mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}.

P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343.

Group the like terms... 1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343.

\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1} > 1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343.

Now for 1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}.

\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}.

Now for 33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}.

Thus we then get \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343.

Now we want to combine fractions. \frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}.

\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2

\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}

\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}.

Expand 81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac.

-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1.

\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}

Therefore P=\frac{-297d+378}{\left(d-1\right)^2}+343.

Hope this helps!

5 0
4 years ago
Please Help! (10 POINTS)
Sliva [168]
The initial measure is:
 m1 = 28 ft
 The real measure must be:
 m2 = 14 ft
 The correct procedure to determine the scale factor is:
 k = m1 / m2&#10;&#10;k = 28/14
 Dividing both numbers between 14 we have:
 k = (28/14) / (14/14)
 Rewriting:
 k = 2/1&#10;&#10;k = 2
 Answer:
 
Zelie should have divided both numbers by 14.
5 0
4 years ago
Read 2 more answers
A gym owner has some purple weights and some white weights. The purple weights are 10 kilograms each, and the white weights are
snow_lady [41]

Answer: 3,9

Step-by-step explanation:

Given

A gym owner has purple and white weights that measure 10 kg and 6 kg each

respectively.

They weigh 84 kg in total

suppose there x purple and y white weights

\therefore x+y=12\quad \ldots(i)\\\Rightarrow 10x+6y=84\quad \ldots(ii)\\\\\text{Solving }(i)\ \text{and }(ii)\\\\

x=3,\ y=9

Thus, there are 3 purple and 9 white weights.

8 0
3 years ago
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