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2 years ago

What is the range for the graph below?

Mathematics

1 answer:

mario62 [17]2 years ago

6 0

Answer:

All real numbers except -3

Step-by-step explanation:

Range refers to numbers on the y-axis.

Domain refers to numbers on the x-axis.

Both of those lines will never reach -3, just come extremely close to it.

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What is the value of x in this triangle?

gizmo_the_mogwai [7]

Answer:

x = 33.49°

Step-by-step explanation:

The relevant trig ratio is the sine:

Sin = Opposite/Hypotenuse

siin(x) = 32/58

The angle measure is found using the inverse function:

x = arcsin(32/58) ≈ 33.49°

8 0

2 years ago

If you drive 23 miles south, make a turn and drive 39 miles east, how far are you, in a straight line, from

Rom4ik [11]

Answer:

45.27 miles

Step-by-step explanation:

Pythagorean theorem

23^2 + 39^2 = d ^ 2

d = 45.27 miles

6 0

1 year ago

A 14-foot ladder is leaning against the side of a house.The angle the ladder makes with the ground is 50 degrees. How far is the

VashaNatasha [74]

Answer:

9 ft

Step-by-step explanation:

You can use cosine to solve this, which is adjacent over hypotenuse. The ladder forms a triangle with the house, and the side you need to solve for is adjacent to the angle. You can set up the equation:

cos(50)=x/14

Then you multiply 14 by cos(50) and get almost 9 ft (rounding to the nearest tenth of a foot).

6 0

3 years ago

What is the slope of the line through (-7.-8) and (0.4)?

Dmitry_Shevchenko [17]

Use the equation (x2-x1)/(y2-y1)

Substitute the value of x and y

7 0

2 years ago

the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?

madam [21]

Answer:

About 0.6548 grams will be remaining.

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

$f(t)=a(r)^t$

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

$\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}$

One year has 365 days.

Therefore, the amount remaining after one year will be:

$\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548$

About 0.6548 grams will be remaining.

Alternatively, we can use the standard exponential growth/decay function modeled by:

$f(t)=Ce^{kt}$

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

$255=510e^{38k}$

Solving for k:

$\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)$

Thus, our function is:

$f(t)=510e^{t\ln(.5)/38}$

Then after one year or 365 days, the amount remaining will be about:

$f(365)=510e^{365\ln(.5)/38}\approx 0.6548$

5 0

2 years ago