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4 years ago

What is the value of the expression? 458+56−134

Mathematics

2 answers:

Fed [463]4 years ago

6 0

Answer:

380

Step-by-step explanation:

Order of operations!

(458+56)-134

(514)-134

514-134

380

done!

DaniilM [7]4 years ago

6 0

Answer:

3.8%

Step-by-step explanation:

After evaluating the expression you get 380. You take 380, and turn it into a percentage, move the decimal to the left twice. Giving you 3.8%

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The function f(x) is a cubic function and the zeros of f(x) are -6, -3 and 1. The

jarptica [38.1K]

Answer:

$P(x)=-5x^3-40x^2-45x+90$

Step-by-step explanation:

Equation of a Polynomial

Given the roots x1, x2, and x3 of a cubic polynomial, the equation can be written as:

$P(x)=a(x-x1)(x-x2)(x-x3)$

Where a is the leading coefficient.

We know the three roots of the polynomial -6, -3, and 1, thus:

$P(x)=a(x+6)(x+3)(x-1)$

Since the y-intercept of the polynomial is y=90 when x=0:

90=a(0+6)(0+3)(0-1)

90=a(6)(3)(-1)=-18a

Thus

a = 90/(-18) = -5

The polynomial is:

$P(x)=-5(x+6)(x+3)(x-1)$

We must write it in standard form, so we have to multiply all of the factors as follows:

$P(x)=-5(x^2+6x+3x+18)(x-1)$

$P(x)=-5(x^2+9x+18)(x-1)$

$P(x)=-5(x^3-x^2+9x^2-9x+18x-18)$

$P(x)=-5(x^3+8x^2+9x-18)$

$\boxed{P(x)=-5x^3-40x^2-45x+90}$

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3 years ago

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3 years ago

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3 years ago

Read 2 more answers

The concentration of dichlorodiphenyltrichloroethane (DDT - an infamous pesticide) in Lake Michigan has been declining exponenti

mojhsa [17]

Answer: A) $y=13e^{-0.1359t}$

B) H = 5.10

C) Yes

Step-by-step explanation: Exponential Decay function is a model that describes the reducing of an amount by a constant rate over time. Generally, it is written in the form: $y(t)=Ce^{rt}$

A) C is initial quantity, in this case, the initial concentration of DDT. To determine r, using the data given:

$y(t)=Ce^{rt}$

$2.22=13e^{13r}$

$e^{13r}=0.1708$

Using a natural logarithm property called power rule:

$13r=ln(0.1708)$

$r=\frac{ln(0.1708)}{13}$

$r=-0.1359$

The decay function for concentration of DDT through the years is $y(t)=13e^{-0.1359t}$

B) The value of H is calculated by $y=C(0.5)^{\frac{t}{H} }$

$2.22=13(0.5)^{\frac{13}{H} }$

$(0.5)^{\frac{13}{H} }=0.1708$

Again, using power rule for logarithm:

$\frac{13}{H} log(0.5)=log(0.1708)$

$\frac{13}{H} =\frac{log(0.1708)}{log(0.5)}$

$\frac{13}{H} =2.55$

H = 5.10

Constant H in the half-life formula is H=5.10

C) Using model $y(t)=13e^{-0.1359t}$ to determine concentration of DDT in 1995:

$y(24)=13e^{-0.1359.24}$

y(24) = 0.5

By 1995, the concentration of DDT is 0.5 ppm, so using this model is possible to reduce such amount and more of DDT.

8 0

3 years ago