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4 years ago

Determine if the equation is linear 2x y=4,

1 answer:

4 0

Determine whether each equation is a linear equation. ... –2x – 3 = y .... linear. 16. 4y. 2. + 9 = –4. SOLUTION: Since y is squared, the equationcannot be written in standard form.

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The variable $x$ varies directly as the square of $y$, and $y$ varies directly as the cube of $z$. If $x$ equals $-16$ when $z$

Kisachek [45]

If $x$ varies directly as $y^2$ , then there is some constant $a$ for which

$x=ay^2$

Similarly, there is some constant $b$ such that

$y=bz^3$

Given that $x=-16$ when $z=2$ , we have

$\begin{cases}-16=ay^2\\y=8b\end{cases}\implies-16=a(8b)^2\implies ab^2=-\dfrac14$

Now when $z=\frac12$ , we get

$\begin{cases}x=ay^2\\y=\frac b8\end{cases}\implies x=a\left(\dfrac b8\right)^2=\dfrac{ab^2}{64}=\dfrac{-\frac14}{64}=\boxed{-\dfrac1{256}}$

6 0

4 years ago

A bag contains 11 red chips and 12 blue chips. Two chips are selected randomly without replacement from the bag. What is the pro

Elena-2011 [213]

Answer:

11/23 or 0.478

Step-by-step explanation:

1. Find the probability of selecting 2 red chips:

11/23 * 10/22 = 5/23

2. Find the probability of selecting 2 blue chips:

12/23 * 11/22 = 6/23

3. Add the two probabilities:

5/23 + 6/23 = 11/23 = 0.478 (rounded to 3 decimal places)

hope this helps! <3

4 0

2 years ago

A standard deck of cards contains 52 cards. one card is selected from the deck. ?(a) compute the probability of randomly selecti

Rama09 [41]

Let
P(h)= probability of selecting heart = 13/52= 1/4
similarly
P(s)= probability of selecting spade =1/4
P(c)= probability of selecting club =1/4
P(d)= probability of selecting diamond =1/4

a)

P(h or s)= P(h)+P(s)= 1/4+1/4=2/4=1/2=0.5

b)

P(h or s or c)=P(h)+P(s)+P(c)=1/4+1/4+1/4=3/4=0.75

c)

P(2 or a diamond)=n(2heart, 2spade,2club,13 diamonds)/total number of cards

=(1+1+1+13)/52=16/52=4/13=0.3

or we could solve it as follows:

P(2 or a diamond)=P(2)+P(diamond)-P(2diamond)=4/52 + 13/52 + 1/52 = (4+13-1)/52=16/52=0.3

3 0

3 years ago

Help me plz help me plz

Lisa [10]

Answer:

A: 1 crackers = 95/4 cals

B: mass= 16/4

c: 15/4

4 0

3 years ago

Read 2 more answers

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