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slava [35]
3 years ago
10

If a line crosses the y-axis at (0,8) and has a slope of -1/8,what is the equation of the line?

Mathematics
2 answers:
IceJOKER [234]3 years ago
8 0

Answer:

d

Step-by-step explanation:

Zina [86]3 years ago
6 0
The answer is D because of the formula "y=mx+b"
the slope is always connected to the x, and "b" is the y axis point of the two coordinate pairs, so it would be minus eight if the coordinate pairs were (0, -8).
y = mx + b
m= slope
b= y axis point of the two coordinate pairs
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In September, Jerry read for 2/5 of an hour every day for 20 days. How many
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Dmitry_Shevchenko [17]

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Read 2 more answers
Consider the solid S described below. The base of S is the triangular region with vertices (0, 0), (4, 0), and (0, 4). Cross-sec
alisha [4.7K]

Answer:

Volume = \frac{64}{3}

Step-by-step explanation:

Given - Consider the solid S described below. The base of S is the triangular region with vertices (0, 0), (4, 0), and (0, 4). Cross-sections perpendicular to the x-axis are squares.

To find - Find the volume V of this solid.

Solution -

Given that,

The equation of the line with both x-intercept and y-intercept as 4 is -

\frac{x}{4} + \frac{y}{4}  = 1

⇒x + y = 4

⇒y = 4 - x

Now,

Volume = \int\limits^a_b {A(x)} \, dx

where

A(x) is the area of general cross-section.

It is given that,

Cross-sections perpendicular to the x-axis are squares.

So,

A(x) = (4 - x)²

As solid lies between x = 0 and x = 4

So,

The Volume becomes

Volume = \int\limits^4_0 {(4 - x)^{2} } \, dx

             = \int\limits^4_0 {[(4)^{2}  + (x)^{2} - 8x] } \, dx

             = \int\limits^4_0 {[16  + x^{2} - 8x] } \, dx

             = {[16 x  + \frac{x^{3}}{3}  - \frac{8x^{2} }{2} ] } ^4_0

             = {[16(4 - 0)  + \frac{4^{3}}{3} - \frac{0^{3}}{3}  - 4 [4^{2} - 0^{2}]   ] }

             = {[16(4)  + \frac{64}{3} - 0  - 4 [16 - 0]   ] }

             = {[64  + \frac{64}{3}  - 64  ] }

             = \frac{64}{3}

⇒Volume = \frac{64}{3}

7 0
2 years ago
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