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3 months ago

What are the domain range and asymptote of h(x)=(1.4)^x+5

Mathematics

2 answers:

Zina [12.3K]3 months ago

8 0

domain: {x | x is a real number}; range: {y | y > 5}; asymptote: y = 5

Inessa [12.5K]3 months ago

6 0

The range consists of all the valid y values, starting from 5.

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The area of an extra large circular pizza from Gambino's Pizzeria is 484\pi \text{ cm}^2484π cm 2 484, pi, space, c, m, start su

zzz [12365]

Answer:

The diameter of a large pizza from Gambino's Pizzeria is $44\ cm$

Step-by-step explanation:

We know that

The area of a circle (circular pizza) is calculated as

$A=\pi r^{2}$

we have

$A=484\pi\ cm^{2}$

substitute values into the equation and resolve for the radius r

$484\pi=\pi r^{2}$

Simplify

$484=r^{2}$

$r=22\ cm$

Determine the diameter

Remember, the diameter is double the radius

$D=2(22)=44\ cm$

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2 months ago

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Complete the formal proof of (~Q→~R)v(R&~Q) from no premises. The empty premise line is not numbered. Hint: there are longer

tester [12383]

Answer:

Step-by-step breakdown:

You may consider a structure like the following while keeping in mind your specific notation.

1. | ~( (~Q ->~R) v (R & ~Q) ) Assume

2. | | ~(~Q ->~R) Assume

3. | | | ~(R & ~Q) Assume

4. | | | ~R v Q 3, De Morgan

5. | | | ~Q -> ~R 4 Material implication

6. | | | # 2, 5 Results in a contradiction

7. | | R & ~Q 3-6 indirect proof outcome

8. | ~(~Q ->~R) -> (R & ~Q ) 2-7 Discharging the conditional proof

9. | (~Q ->~R) v (R & ~Q) 8 Material implication derived

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2 months ago

Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inche

tester [12383]

Response:

The height of the water when it reaches 4 inches is $\frac{3}{8\times \pi} inch/s$ .

Detailed Explanation:

Flow rate of water from the cone = R= $\frac{3}{2} inch^3/s$

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

$\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}$

r = h/2

Volume of the cone = $V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi r^2h$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}$

$\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}$

$\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

h = 4 inches

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2$

$\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s$

The height of the water when it is 4 inches deep is $\frac{3}{8\times \pi} inch/s$ .

6 0

3 months ago