All categories

2 years ago

Y=-1/3-5 Graph this line using the slope and y-intercept.

Mathematics

1 answer:

ioda2 years ago

8 0

Big old buckaroo faxual

You might be interested in

Nate solves the equation _ x 5=15 by writing and solving 15/5=_. Explain why Nates method works.

Crazy boy [7]

This is because Nate is isolating the variable. For example, in the equation 10x=40, you would divide both sides by 10 in order to get the variable by itself, just as Nate did. Nate is also performing the inverse operation, which allows for one to undo/unravel the question to find the answer.

3 0

2 years ago

Solve the formula 180=a+b+c for a

navik [9.2K]

Answer:

a = 180 - b - c

Step-by-step explanation:

given 180 = a + b + c or

a + b + c = 180 ( isolate a by subtracting b from both sides then c )

subtract b from both sides

a + c = 180 - b

subtract c from both sides

a = 180 - b - c

8 0

2 years ago

To which interval would we restrict f(x) = cos (x-π/4) so that f(x) is invertible?

nadezda [96]

Answer:

The range of x values for which y is unique is 2·π

Step-by-step explanation:

For a function j: X → Y to be invertible, we have that for every y in Y, there is associated only one x which is an element of x

Hence, f(x) = cos(x - π/4) gives

the x intercept at two penultimate points of the graph of cos(x - π/4) are;

x = 2.36, and x = 8.64

x = 3/4·π, and x = 2.75·π = $2\tfrac{3}{4}\cdot \pi$

Hence the range of x values for which y is unique is presented as follows

$2\tfrac{3}{4}\cdot \pi - \frac{3}{4}\cdot \pi = 2}\cdot \pi$

The range of x values for which y is unique = 2·π.

4 0

2 years ago

Look at the graph

Ira Lisetskai [31]

y increases as x increases

3 0

2 years ago

Read 2 more answers

The z-score associated with 95% is 1.96. If the sample mean is 200 and the standard deviation is 30, find the upper limit of the

ziro4ka [17]

Answer:

The upper limit of the 95% confidence interval is:

C.I_u = 200 + (58.8/ $\sqrt{n}$ )

Step-by-step explanation:

The formula is given as:

C.I = μ ± Z*σ/ $\sqrt{n}$

The upper limit => C.I_u = μ + Z*σ/ $\sqrt{n}$

The lower limit => C.I_l = μ - Z*σ/ $\sqrt{n}$

The sample size (n) is not stated in the question. Hence, we calculate the upper limit with respect to n.

The upper limit => C.I_u = 200 + 1.96*(30/ $\sqrt{n}$ )

= 200 + (1.96*30)/ $\sqrt{n}$

= 200 + 58.8/ $\sqrt{n}$

5 0

2 years ago