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4 years ago

(x-2)times (x^2+8x-1)

Mathematics

1 answer:

DerKrebs [107]4 years ago

6 0

Answer:

x^3+6x^2-17x+2

Step-by-step explanation:

(x-2)*(x^2+8x-1)

=x(x^2+8x-1)-2(x^2+8x-1)

=x^3+8x^2-x-2x^2-16x+2

collect like terms

= x^3+6x^2-17x+2

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Log8 +logy =1 Plz help

Alexeev081 [22]

Answer:

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Step-by-step explanation:

log8 + logy = 1

logy =1 - log8

logy = 0.0969

y = 1.25

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4 years ago

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3 years ago

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Solve for y: x-0.25y=1.5

STatiana [176]

To find value of y in times o x we have to transfrom this equation.
LEts do this
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3 years ago

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A line passes through the point (-9,3) and has a slope of -2/3?

saveliy_v [14]

The point-slope form:

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

We have the slope m = -2/3 and the point (-9, 3). Substitute:

$y-3=-\dfrac{2}{3}(x-(-9))$

$y-3=-\dfrac{2}{3}(x+9)$ use distributive property

$y-3=-\dfrac{2}{3}x-\dfrac{2}{3}\cdot9$

$y-3=-\dfrac{2}{3}x-(2)(3)$

$y-3=-\dfrac{2}{3}x-6$ add 3 to both sides

$\boxed{y=-\dfrac{2}{3}x-3}$

5 0

3 years ago

The population of a town with a 20162016 population of 87 comma 00087,000 grows at a rate of 1.91.9% per year. a. Find the rate

SCORPION-xisa [38]

Answer:

Given,

The initial population, P = 87,000,

Annual rate of increasing, r = 1.9 %,

a) Thus, the population after t years,

$A=P(1+\frac{r}{100})^t$

$A=87000(1+\frac{1.9}{100})^{t}$

$=87000(1+0.019)^t$

$=87000(1.019)^t$

$87000(1.019)^t=87000e^{kt}$

Where, k is the rate constant,

By comparing,

$\implies k=log(1.019)=0.00817418400643\approx 0.00817$

Hence, the approximate value of k is 0.00817.

And, the exponential growth function would be,

$A=87000e^{0.00817t}$

b) If A = 145,000,

$145000=87000(1.019)^t$

By using graphing calculator,

We get,

$t=27.14\approx 27$

The year after 27 years from 2016 is 2043.

Hence, in 2043 the population reach 145,000.

6 0

4 years ago