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2 years ago

Help please I need this as soon as posible

Mathematics

2 answers:

Romashka [77]2 years ago

3 0

Answer:

y=2x+4

Step-by-step explanation:

the y intercept is 4, so you plug that into the equation, and since the line goes up 4 then over 2 so when you put rise over run, the slope is positive 2, which you can verify since the line is going up

*note, good job on figuring it out on your own. i know that you probably second guess your self, but 8 times out of 10, your instincts are correct

if you have any questions, leave them in the comments and i will try to answer them, if this helped, pls give brainliest

Elan Coil [88]2 years ago

3 0

Y=2x+4 ???????????????

You might be interested in

29. The function f(t) =-16t^2 + 35 gives the height, f (in feet) of a ball t seconds after being dropped from a height of 35 fee

Nostrana [21]

The time taken for the ball to fall a distance of 10feet is 1.25secs

Given the function modeled by the height expressed as:

f(t) =-16t² + 35

In order to determine the time taken by the all to fall a distance of 10 feet, hence;

f(t) = 10

10 = -16t² + 35

-16t² = 10 - 35

-16t² = -25

-t² = -25/16

t² = 25/16

t = √25/16

t = 5/4

t = 1.25secs

Hence the time taken for the ball to fall a distance of 10feet is 1.25secs

Learn more here: brainly.com/question/13881652

8 0

3 years ago

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

PLEASE HELP!!!!!!!!

Korolek [52]

Answer:

2π - 4√5

(= -2.66 to nearest hundredth).

Step-by-step explanation:

√(2π - 4√5)^2

= 2π - 4√5 (by the definition of a square root)

If you want an approximate value it is

-2.66 to the nearest hundredth

7 0

2 years ago

What is the average (arithmetic mean) of the first 90 measurements in a certain list of 92 measurements? (1) the average of the

marishachu [46]

The sum of all the measurements is just the average times the number.

$\bar x = \frac{1}{n} \sum_{i=1}^n x_i$

$\sum_{i=1}^n x_i = n \bar x$

So if the average of all 92 is 7 the sum of those is $92 \times 7$

The average of the last two is 7.2 so their sum is $2 \times 7.2$

That means the sum of the first 90 is $92 \times 7 - 2 \times 7.2$

so the average of the first 90 is

$\dfrac{92 \times 7 - 2 \times 7.2}{90} = 6.99556$ cm

7 0

3 years ago

(Question Is In The Picture, Kinda Bad)

11111nata11111 [884]

Answer:

72 cupcakes

Step-by-step explanation:

Morgan can make 4 cupcakes with one cup of flour. She has 18 cups of flour and suppose she can make x cupcakes, then

1 cup of flour - 4 cupcakes

18 cups of flour - x cupcakes

Write a proportion:

$\dfrac{1}{18}=\dfrac{4}{x}$

Cross multiply:

$1\cdot x=18\cdot 4\\ \\x=72\ cupcakes$

5 0

3 years ago

Help please I need this as soon as posible ​

Help please I need this as soon as posible