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3 years ago

Which polynomial is written in ascending order? t2 − t 4 t2 4 − t 4 t2 − t 4 − t t2

2 answers:

4 0

In order to solve this, you have to look at the order for each of the variable terms. In this case t is your variablr
4t0=4 exponent is 0. −t1=−t exponent is 1. t2 exponent is 2. So you order them by exponents from least to greatest (ascending). 4−t+t2

SOVA2 [1]3 years ago

4 0

Answer:

I had this question on a test and I got it right with 4 − t + t2

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30 people liked Brand X. This is 20% of the people in a survey.

Yuki888 [10]

A) 150 people
b) 80%
c) 120 people

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3 years ago

Ashley's Internet service is terribly unreliable. In fact, on any given day, there is a 15% chance that her Internet connection

stepan [7]

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Step-by-step explanation:

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3 years ago

The shower block on the campsite has an area of 144 m2 The length of the shower block is 24 m. What is the width? m

skad [1K]

Area= length*width

Plug in what you know.

144= 24*width

Divide both sides by 24.

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I hope this helps!
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4 0

3 years ago

A town has a population of 12000 and grows at 3.5% every year. What will be the population after 7 years, to the nearest whole n

Setler79 [48]

Answer:

15,267

Step-by-step explanation:

To solve this problem, lets use the exponential growth formula which is: $A = P(1+r)^t$

A = total amount

P = original amount

r = growth rate (decimal)

t = years

Before we plug in the values, don't forget to change 3.5% to its decimal form.

3.5% -> $\frac{3.5}{100}$ -> 0.035

Lets plug in the values:

$A=12,000(1+0.035)^7$

$A=15,267$

The population after 7 years will be 15,267.

8 0

3 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

8 0

3 years ago