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igor_vitrenko [27]

3 years ago

11

What is the sign of y = £? y Choose 1 answer: A Positive B Negative Zero

1 answer:

KATRIN_1 [288]3 years ago

4 0

Answer:

I'm pretty sure is

Positive b

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Solve the system of linear equations by substitution.<br> 2x-y=3<br> x+5y=14

adoni [48]

Let 2x – y = 3 ——— equation 1

Let x + 5y = 14 ——— equation 2

Making x the subject in eqn 1, = x = y + 3 / 2 ——— eqn 3

• Put eqn 3 in eqn 2

(y + 3 / 2) + 5y = 14

6y = 14 – 3/2

6y = 25/2

y = 25/12

• put y = 25/12 in eqn 3

x = (25/12 + 3/2)

x = 43/12

4 0

3 years ago

A right-angled triangle has sides which are 5 inches and 12 inches. How long is the triangle's hypotenuse?

Andrew [12]

Answer:

13 inches

Step-by-step explanation:

Use the Pythagorean Theorem:

a^2 + b^2 = c^2

(5)^2 + (12)^2 = c^2

c^2 = 169 in^2

c = 13 inches

4 0

2 years ago

Joshua randomly selects a vehicle to test drive. There are 12 cars, 9 trucks, and 4 vans on the lot. What is the probability Jos

eimsori [14]

Step-by-step explanation:

<em>The</em><em> </em><em>number</em><em> </em><em><u>of</u></em><em><u> </u></em><em><u>vehicle</u></em><em><u>s</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>2</u></em><em><u>5</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>no</u></em><em><u>.</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>truck </u></em><em><u>is</u></em><em><u> </u></em><em><u>9</u></em><em><u> </u></em><em><u>so</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>probability </u></em><em><u>is</u></em><em><u> </u></em><em><u>9</u></em><em><u>/</u></em><em><u>2</u></em><em><u>5</u></em>

4 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

Need a correct answer please

zloy xaker [14]

Answer x=14 y=1

Step-by-step explanation:

3 0

3 years ago