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Alona [7]
4 years ago
8

What does y=mx+b means

Mathematics
2 answers:
soldier1979 [14.2K]4 years ago
7 0

Answer:

y=mx+b is "Slope-Intercept Form"

In an equation (when the equation is written as "y = mx + b"), the slope is the number ("m") that is multiplied by the x, and "b" is the y-intercept (point where the line crosses the vertical y-axis). This form of the line equation is called the "slope-intercept form". Hope this helps! ^-^


marusya05 [52]4 years ago
7 0

Slope Intercept Form!
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A cylinder has a diameter of 40 feet and a height of 32 feet. What is the surface area of the cylinder? Express the answer in
mafiozo [28]

Answer:

6524.51 ft^2

Step-by-step explanation:

The surface area of a cylinder is given as;

SA = 2\pi rh + 2\pi r^2

where r = radius

h = height

The cylinder has a height of 32 feet and a diameter of 40 feet. This means that the radius is:

40 / 2 = 20 feet

The surface area is therefore:

SA = (2*\pi* 20 * 32) + (2*\pi *20^2)\\\\SA = 4021.24 + 2513.27\\\\SA = 6524.51 ft^2

The surface area of the cylinder is 6524.51 ft^2.

5 0
3 years ago
100 POINTS+ BRAINLYEST❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
Karo-lina-s [1.5K]

Step-by-step explanation:

The vertex form of a parabola can be written as

y =  {a(x - h)}^{2}  + k

where (h, k) are the coordinates of the parabola's vertex. The vertices of the functions are as follows:

a) (3, 5)

b) (-7, 3)

c) (4, 0)

d) (0, -1)

e) (-1, -5)

f) (-1, -5)

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3 years ago
Question
dezoksy [38]

Answer:

the height of the triangle is 47 inches

4 0
3 years ago
Find the slope of the line that passes through (-34, 72) and (31, 27).
ZanzabumX [31]
The answer i got is -9/13
8 0
3 years ago
Read 2 more answers
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
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