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2 years ago

What is the value of x A. 100 B.50 C.25 D.35

Mathematics

2 answers:

PSYCHO15rus [73]2 years ago

5 0

Answer:

Step-by-step explanation:

I believe this is correct

mestny [16]2 years ago

4 0

D.35. I’m pretty sure this is the correct answer as well

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NEED THIS QUICK, WILL GIVE BRAINLIEST

grin007 [14]

Answer:

The slope is 10.

Step-by-step explanation:

30/3 = 10

50/5 = 10

70/7 = 10

90/9 = 10

The common factor is 10 which is the slope.

4 0

3 years ago

Homework ree

umka21 [38]

Answer:

325 miles and 77 dollars

Step-by-step explanation:

First, create equations. Let x equal the miles driven. The first plan expression is 38 + 0.12x and the second plan expression is 51 + 0.08x. Setting these equal and solve for x: 38 + 0.12x = 51 + 0.08x -> 0.04x = 13 -> x = 325 miles.

The cost is found by substituting it into any of the expressions: 51 + 0.08(325) = 77 or 38 + 0.12(325) = 77

6 0

3 years ago

the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a

Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - First derivative of mass in time, measured in miligrams per year.

$\tau$ - Time constant, measured in years.

$m$ - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

$\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt$

$\ln m = -\frac{1}{\tau} + C$

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of isotope, measured in miligrams.

$t$ - Time, measured in years.

And time is cleared within the equation:

$t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]$

Then, time constant can be found as a function of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $t_{1/2} = 5730\,yr$ and $\frac{m(t)}{m_{o}} = 0.35$ , then:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.35$

$t \approx 8678.505\,yr$

The wood was cut approximately 8679 years ago.

5 0

3 years ago

A garden hose emits 9 quarts of water in 6 seconds how much did the water emit in 10seconds

jarptica [38.1K]

Answer:

In 10 seconds, the garden hose will emit 15 quarts of water.

Step-by-step explanation:

The amount of water emitted by the garden hose over time can be expressed as a ratio: 9/6, or 9 quarts of water for every 6 seconds of time. We can then simplify this ratio to 3/2, or 3 quarts of water for every 2 seconds of time. Since the ratio will remain constant, or the same, over time, we can set up an equivalent ratio, or fraction to find the amount of water emitted in 10 seconds: 3/2 = x/10. We look at the denominators and see that 2 x 5 = 10. In order to make the ratios equivalent, we would also multiply the numerator by 5: 3 x 5 = 15, which gives us the amount of water emitted in 10 seconds.

7 0

2 years ago

(8Q) Tell whether the function exhibits damped oscillation. If it does, identify the damping factor and tell whether the damping

dezoksy [38]

Answer:

Option c.

No damping

Step-by-step explanation:

We can easily solve this question by using a graphing calculator or any plotting tool.

The function is

f(x) = (√11)*cos(3.7x)

Which can be seen in the picture below

We can notice that f(x) is a cosine with maximum amplitude of (√11). Neither this factor nor the multiplication of x by 3.7 serve as a damping factor since they are constants.

f(x) does not present any dampening

5 0

3 years ago

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