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3 years ago

Write the equation of the line that passes through the points (0, -5) and (2, -1).

Mathematics

1 answer:

IRISSAK [1]3 years ago

5 0

Answer:

y = -2x + 3

Step-by-step explanation:

To find the slope, we can use this formula: y2 - y1/x2 - x1

Y2 is the y - coordinate from the second coordinate, so you should be able to figure out what the rest of the letters represent in that formula.

So we would plug in the coordinates and find the slope.

-1 - (-5)/2 - 0

4/-2

slope = m = -2

We know the slope and coordinates, so we can plug those into the equation and find the y intercept, or b.

y = mx + b

-1 = -2(2) + b

-1 = -4 + b

3 = b

Now we know the slope and y intercept so we can put them into the equation to find the final equation, which is: y = -2x + 3

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Which situation can be represented by the expression <br><br> 5−8

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3 years ago

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Divide by 4 : x =.68

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3 years ago

Find the surface area of the triangular prism (above) using its net (below).

r-ruslan [8.4K]

Answer:

96 square units

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.

My answer:

The length of the large rectangle is: 5
The width of the large rectangle is: 7

=> Area of the large rectangle = length × width = 7*5 = 35 square units

The length of the middle rectangle is: 7
The width of the middle rectangle is: 4

=> Area of the middle rectangle = length × width = 7*4 = 28 square units

The length of the small rectangle is: 7
The width of the small rectangle is: 3

=> Area of the small rectangle = length × width = 7*3 = 21 square units

Area of top and the bottom triangles =2* $\frac{1}{2}bh$ $= 4*3 =12$

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3 years ago

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Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt

sveta [45]

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

----

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}

= 26/15.

----

For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).

So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

= 5/π.

Therefore,

∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.

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3 years ago