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3 years ago

How do i calculate the slope of the line between (-2, -4) and (-10, 5)

Mathematics

2 answers:

Goryan [66]3 years ago

8 0

Look at the image. I need 20 characters lol

luda_lava [24]3 years ago

6 0

Answer:

-9/8

Step-by-step explanation:

To solve this you do the change in y over the change in x. so in this case it would be (-4)-(5)/(-2)-(-10). Which is -9/8 which is the slope.

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Katalin drove 300 miles on her vacation. She drove an average of 1.9 times faster on the second 150 miles of her trip than she d

viktelen [127]

300=x+1.9x
I got this by dividing the problem into two parts x[the speed of the first half] plus 1.9x [the speed of the second half increased by 1.9] both of those together should give you 300.
I went further and solved this...If you placed 103.5 into this problem for x you should get 300.15!

8 0

3 years ago

Solve.<br> 3(5z + 7) = 33

Ivenika [448]

Answer:

z = 4/5

Step-by-step explanation:

3(5z + 7) = 33

Divide each side by 3

3/3(5z + 7) = 33/3

5z+7 = 11

Subtract 7 from each side

5z+7-7 = 11-7

5z = 4

Divide by 5

5z/5 = 4/5

z = 4/5

3 0

3 years ago

Read 2 more answers

Olivia bought a dress for $120. The tax rate is 9%. What is the total amount Olivia paid for the dress?

elixir [45]

I think she will pay $130.80

7 0

2 years ago

Read 2 more answers

The illuminance of a surface varies inversely with the square of its distance from the light source.If the illuminance of a surf

VashaNatasha [74]

Answer:

Distance between the surface and source of light will be increased by 6 meters.

Step-by-step explanation:

The illuminance of a surface varies inversely with the square of ts distance from the light source.

Let illuminance of the surface = x lumens per square meter

and distance from a light source = y meter.

Now x ∝ $\frac{1}{y^{2} }$

Or $x=\frac{k}{y^{2} }$ [k = proportionality constant]

Now we will find the value of k.

k = xy²

k = 120×(6)²

k = 4320

We have to calculate the distance of the source if illuminance of the surface is 30 lumens per square meter.

$30=\frac{4320}{y^{2}}$

y² = $\frac{4320}{30}$

y² = 144

y = √144 = 12 meters

So the source of the light will be shifted away from the surface = 12 - 6 = 6 meters.

6 0

2 years ago

The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the

sweet-ann [11.9K]

Answer:

24 cm wide by 36 cm high

Step-by-step explanation:

The poster with the smallest area will have an aspect ratio that makes the margin dimensions the same percentage of overall dimension in each direction.

Since the ratio of margin widths is 6:4 = 3:2, the poster and printed area will have an aspect ratio of 3:2. That is, the width is ...

width of printed area = √(2/3·384 cm²) = 16 cm

Then the width of the poster is ...

width = left margin + printed width + right margin = 4cm + 16 cm + 4 cm

width = 24 cm

The height is 3/2 times that, or 36 cm.

The smallest poster with the required dimensions is 24 cm wide by 36 cm tall.

_____

If you need to see the calculus problem, consider the printed area width to be x. Then the printed height is 384/x and the overall dimensions are ...

(x + 8) by (384/x + 12)

We want to minimize the area, which is the product of these dimensions:

a = (x +8)(384/x +12) = 384 +12x +3072/x +96

a = 12x + 3072/x +480

This is a minimum where its derivative is zero.

a' = 12 -3072/x^2 = 0

a' = 1 -256/x^2 = 0 . . . . . . divide by 12; true when x^2 = 256

This has solutions x=±16, of which the only useful solution is x=16.

7 0

3 years ago