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3 years ago

11

point x is at 2/3 on a number line. on the same number line, point y is the same distance from 0 as point x, but has a numerator

of 8. what is the denominator of the fraction at point y?

1 answer:

lapo4ka [179]3 years ago

7 0

The denominator of the fraction at point Y would be 24.

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Which function is equivalent to the inverse of y=sec(x)

kipiarov [429]

From the identity:

$sec(x)= \frac{1}{cos(x)}$

$f(x)=sec(x)= \frac{1}{cos(x)}$

the inverse of f is g such that f(g(x))=x,

we must find g(x), such that $\frac{1}{cos[g(x)]}=x$

thus, $cos[g(x)]= \frac{1}{x}$

$g(x)=cos^{-1} (\frac{1}{x})$

Answer: b. g(x)=cos^-1(1/x)

6 0

3 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

3 years ago

N–2=10n+4/2 what is the value of n need help asap

MakcuM [25]

N - 2 = 10n + 4/2

-2 = 9n + 4/2

-4/4 = 9n

-1 = 9n

-9 = n

N is -9.

If this answer is right please leave a thanks and a rating. :)

4 0

3 years ago

What is the ratio equivalent to 0.l

NeTakaya

1 / 10 is the easiest answer. Remember ratios are just percentages. It could also be 2/20, 3/30, as long as it is 10%.

7 0

3 years ago

Identify the coordinates of the vertices of each figure after the similarity transformation.

bija089 [108]

Answer:

After translation:

Translation: (x, y) → (x-6, y-8)

W'X'Y'Z' = W'(-10, 0), X'(-10, -8), Y'(-18, -4), Z'(-14, 8)

After dilation:

Dilation: (x, y) → (2x, 2y), centered at (0, 0)

W''X''Y''Z'' = W'' (-20, 0), X'' (-8, 0) , Y'' (-24, 8), Z'' (-16, 32)

Step-by-step explanation:

Given the vertices of the Quadrilateral WXYZ

W(-4, 8)

X(-4, 0)

Y(-12, 4)

Z(-8, 16)

As the rule of translation is given by

Translation: (x, y) → (x-6, y-8)

So, after the translation the vertices of Quadrilateral WXYZ will be translated as:

W(-4, 8) → (x-6, y-8) ⇒ (-4-6, 8-8) = W'(-10, 0)

X(-4, 0) → (x-6, y-8) ⇒ (-4-6, 0-8) = X'(-10, -8)

Y(-12, 4) → (x-6, y-8) ⇒ (-12-6, 4-8) = Y'(-18, -4)

Z(-8, 16) → (x-6, y-8) ⇒ (-8-6, 16-8) = Z'(-14, 8)

Therefore,

W'X'Y'Z' = W'(-10, 0), X'(-10, -8), Y'(-18, -4), Z'(-14, 8)

Next, the rule of dilation is given by

Dilation: (x, y) → (2x, 2y), centered at (0, 0)

W'(-10, 0) → (2x, 2y) ⇒ (2(-10), 2(0)) = W'' (-20, 0)

X'(-4, 0) → (2x, 2y) ⇒ (2(-4), 2(0)) = X'' (-8, 0)

Y'(-12, 4) → (2x, 2y) ⇒ (2(-12), 2(4)) = Y'' (-24, 8)

Z'(-8, 16) → (2x, 2y) ⇒ (2(-8), 2(16)) = Z'' (-16, 32)

Therefore,

W''X''Y''Z'' = W'' (-20, 0), X'' (-8, 0) , Y'' (-24, 8), Z'' (-16, 32)

8 0

3 years ago