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4 years ago

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 $6524.51 ft^2$ .

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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:

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

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

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

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

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3 years ago