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3 years ago

In a proportional relationship why must the graph of the line always go through the origin (0,0)? explain?

Mathematics

1 answer:

lbvjy [14]3 years ago

7 0

Answer:

We know that, two quantities are proportional when they vary directly with each other.

i.e. quantities x and y are proportional when they have the relation y = kx, where k is any constant called constant of proportionality.

Since, the quantities are in relation y = kx.

So, for x = 0 we get that the value of y = 0.

As, the point ( 0,0 ) satisfies this relation i.e. it is a solution of y = kx.

Hence, in proportional relationship, the graph will pass through ( 0,0 ).

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Neko [114]

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3 years ago

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

Use the real zeros to factor f x^4 plus 10^3 minus 20x^2 minus 90x plus 99

Furkat [3]

Answer:

f(x) = (x + 11) (x + 3) (x − 1) (x − 3)

Step-by-step explanation:

f(x) = x⁴ + 10x³ − 20x² − 90x + 99

f(x) is a fourth order polynomial, so it has 4 roots. Use rational root theorem to find possible rational roots.

±99, ±11, ±9, ±3, ±1

By trial and error, the zeros are -11, -3, 1, and 3.

f(x) = (x + 11) (x + 3) (x − 1) (x − 3)

7 0

3 years ago

A line with a slope of -1/3 passes through the point (9,3). What is its equation in point-slope form?

amm1812

The point-slope form:

$y-y_1=m(x-x_1)$

We have:

$(9,\ 3)\to x_1=9,\ y_1=3\\\\m=-\dfrac{1}{3}$

Substitute

$y-3=-\dfrac{1}{3}(x-9)$

7 0

3 years ago

How do you work out a percentile rank of a score of 57

arlik [135]

To do that you'll need the mean and standard deviation of all the scores. Can you provide this info?

For example: Supposing that the mean of these scores were 52 and the standard deviation 3. You'd need to find the "z-score" of 57 in this case.
57 - 52
It is z = ------------ , or z = 5/3, or z = 1.67.
3

Find the area to the left of z = 1.67. Multiply that area by 100% to find the percentile rank of the score 57.

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3 years ago