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3 years ago

Who drew a line with a slope of 3? names are above each graph

Mathematics

1 answer:

hoa [83]3 years ago

4 0

Answer:

Benito did

Step-by-step explanation:

a slope is basically rise over run so u can see that he started at -1.

-1 plus 3 is 2

then the run part is just going right one since its a whole number

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Determine whether the lines are parallel, perpendicular, or neither. 5x + 2y = 14 and y = −5x + 9

valentinak56 [21]

Answer: Neither

Step-by-step explanation:

5x + 2y = 14

2y = -5x + 14

y = -5/2 x + 7; slope m = -5/2

y = –5x + 9 ; slope m = -5.

Parallel lines slopes are equal and perpendicular slopes are opposite and reciprocal.

So both lines are neither parallel nor perpendicular

Answer

C. Neither.

8 0

2 years ago

Employees at Standard Storage are in the ratio of 2 women to 7 men. If there are 490 men employed,

PilotLPTM [1.2K]

Proportion women /men = 2/7 = x/490 7x= 2· 490 :7

X = 2·70 = 140

4 0

2 years ago

One morning, the temperature in Urbana, Illinois, was 15° Fahrenheit. By evening, the temperature had dropped 25° Fahrenheit. Wh

Grace [21]

The temperature started at 15 degrees.

It dropped 25 degrees.

Subtract 25 from 15:

15 - 25 = -10

The temperature was -10 degrees.

5 0

3 years ago

20 POINTS ASAP The graphed line can be expressed by which equation? Math item stem image y+3=32(x+1) y−0=−32(x+1) y−3=32(x−1) y−

storchak [24]

Answer:

The Answer is: y - 3 = 3/2(x - 1)

Step-by-step explanation:

Given Points: (1, 3) and (-3, -3)

Find the slope m:

m = y - y1 / (x - x1)

m = 3 - (-3) / (1 - (-3))

m = 3 + 3 / 1 + 3

m = 6 / 4 = 3/2

Use the point slope form and point (1, 3):

y - y1 = m(x - x1)

y - 3 = 3/2(x - 1)

Hope this helps! Have an Awesome day!! :-)

7 0

3 years ago

Read 2 more answers

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