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3 years ago

What information does a supply schedule provide?

Mathematics

2 answers:

Tju [1.3M]3 years ago

7 0

Answer:

c. It shows the supply for a product at various prices.

Step-by-step explanation:

dimulka [17.4K]3 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

1. P(x)=x^2+2<br> 2. P(x)= 2x +10

ruslelena [56]

Answer:

see explanation

Step-by-step explanation:

Since p(x) = x² + 2 , if x ≤ 4

For x = 4 then this is included in the inequality x ≤ 4

whereas x > 4 does not include x = 4 but values greater than 4

Thus to evaluate x = 4 use p(x) = x² + 2

4 0

4 years ago

Miguel's scooter can travel at a maximum speed of 45 miles per hour. Which inequality models all the speeds, at which Miguel's s

Iteru [2.4K]

Answer:

I can't put the symbol in but less then or equal to 45mph

Step-by-step explanation:

4 0

3 years ago

Which best describes the error in finding the percent increase from 18 to 26?

mrs_skeptik [129]

Answer:

Step-by-step explanation:

The denominator should be the initial value of 18.

8 0

3 years ago

What are the first 3 digits of the square root of Pi?

Tanya [424]

Hello!!

The square root of pi 3.14 is:

$1.77245385$

I am sure you can figure out the main point!

Good luck :)

6 0

3 years ago

Read 2 more answers

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago