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2 years ago

A hyperbola centered at the origin has a vertex at (0, 36) and a focus at (0, 39).

2 answers:

4 0

To solve this problem you must apply the proccedure shown below:
1. You have that the hyperbola <span>has a vertex at (0,36) and a focus at (0,39).
2. Therefore, the equation of the directrices is:

a=36
a^2=1296
c=39

y=a^2/c

3. When you susbtitute the values of a^2 and c into </span>y=a^2/c, you obtain:
<span>
</span>y=a^2/c
<span> y=1296/13

4. When you simplify:

y=432/13

Therefore, the answer is: </span><span>y = ±432/13</span>

Anit [1.1K]2 years ago

4 0

Answer:

The answer is D

Step-by-step explanation:

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Kiley can complete 35 math problems in 1 1/4 hours.How many can she complete in one hour

Lemur [1.5K]

Answer: 28

Step-by-step explanation:

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1 year ago

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Nisha works at a bank. In her till 35% of the notes are £5 notes, 40% are £10 notes and the rest are £20 notes. If all the £20 n

podryga [215]

Answer:

£1290

Step-by-step explanation:

35%+40%= 75%

£600=25% £600÷20=30 25%of till = 30 notes 100%=120 notes

40% of 120 =48notes 48x10=£480

90notes-48notes=42notes

42x£5=£210

£600+£480+£210=£1290

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1 year ago

When an apple is dropped from a tower 256 feet high, the function h(t) =-167 +256 models the height of the

DerKrebs [107]

Answer:

around 1.5 sec

Step-by-step explanation:

basically you wanna figure out at what time is the height=0

since h(t) represents height, set it to 0 then solve for t

i believe you might have forgotten the t in the equation so i assumed it was -167t

0=-167t+256

-256=-167t

t=1.53293413174

around 1.5 seconds after it was dropped

alternatively, you could plug the equation into desmos, replacing h(t) with y and t with x and find the x intercept

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1 year ago

Two professors are applying for grants. Professor Jane has a probability of 0.61 of being funded. Professor Joe has probability

Anna11 [10]

Answer:

a. 0.1647;
b. 0.7153;
c. 0.4453.

Step-by-step explanation:

By definition, if two event $A$ and $B$ are independent, then

$P(A\cap B) = P(A) \cdot P(B)$ . ( $P(A\cap B)$ is the probability that the outcome of both event $A$ and event $B$ are true.)

Since the outcome of these two events are independent,

$\begin{aligned}P(\texttt{Jane} \cap \texttt{Joe}) &= P(\texttt{Jane}) \cdot P(\texttt{Joe})\\ &= 0.61 \times 0.27 \\&= 0.1647\end{aligned}$ .

The logic not operator $\lnot$ or the prime superscript $^{\prime}$ denotes that an event does not happen.

$P(\texttt{Jane}^{\prime}) = 1 - P(\texttt{Jane}) = 1- 0.61 = 0.39$ .

$P(\texttt{Joe}^{\prime}) = 1 - P(\texttt{Joe}) = 1- 0.27 = 0.73$ .

Since the two events $\texttt{Jane}$ and $\texttt{Joe}$ , $\texttt{Jane}^{\prime}$ and $\texttt{Joe}^{\prime}$ are also independent. Probability that neither professor got funded:

$P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = P(\texttt{Jane}^{\prime}) \cdot P(\texttt{Joe}^{\prime}) = 0.39 \times 0.73 = 0.2847$ .

Probability that at least one professor got funded- in other words, it is not true that neither professor got funded:

$P((\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime})^{\prime}) = 1- P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = 0.7153$ .

Similarly, since the two events $\texttt{Jane}$ and $\texttt{Joe}$ , $\texttt{Jane}$ and $\texttt{Joe}^{\prime}$ are also independent. Probability that Jane but not Joe got funded:

$P(\texttt{Jane} \cap (\texttt{Joe}^{\prime})) = P(\texttt{Jane}) \cdot P(\texttt{Joe}^{\prime}) = 0.61 \times 0.73 = 0.4453$ .

4 0

1 year ago

Which equations have the variable term in the equation 6 + 2x = 6x - 9 isolated to one side of the equals sign, and

MrRissso [65]

Answer:

-4x = -3

3 = 4x

7 0

1 year ago

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