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IrinaK [193]
3 years ago
7

2x+3=9 What is x? this is an awesome question

Mathematics
2 answers:
sergejj [24]3 years ago
5 0
X is the variable in the equation. 
2x + 3 = 9
2x = 6
x = 3
artcher [175]3 years ago
5 0
2x+3 = 9
         -3     -3 
2x     = 6

2x/2   = 6/2
x      =  3
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In 2012, the population of a City was 6.89 million. The exponential growth rate was 3.13% per year
MrRa [10]

Answer:

in 2018 it will be 8.09 million

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3 years ago
How many pounds are in 1 ton 1,600 ounces?
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I believe there are 32000 ounces in a ton.
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Which will result in a difference of squares?
otez555 [7]

Expanding the given expressions using Foil:

1)(–7x + 4)(–7x + 4) =7x^{2}-28x-28x+16=49x^{2}-56x+16

2) (–7x + 4)(4 – 7x)=-28x+49x^{2}+16-28x=49x^{2}-56x+16.

3)(–7x + 4)(–7x – 4)=49x^{2}+28x-28x-16=49x^{2}-16.

4)(–7x + 4)(7x – 4)=49x^{2}+28x+28x-16=49x^{2}+56x-16.

The third option that is (–7x + 4)(–7x – 4) is difference of two squares.

5 0
3 years ago
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Tristan is working two summer jobs, making $7 per hour babysitting and $19 per hour lifeguarding. Tristan must earn at least $21
rodikova [14]

Answer: 7bb +19 ll ≥210

Step-by-step explanation:

Hi, to answer this question we have to write an inequality:

The product of the number of hours he works babysitting (bb) and the amount he earns per hour (7); plus The product of the number of hours he works lifeguarding (ll) and the amount he earns per hour 19; must be higher or equal to the amount he must earn this week (210)

Mathematically speaking:

7 bb + 19 ll ≥210

6 0
3 years ago
Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95
Virty [35]

Answer:

L(x)=1+\dfrac{1}{3}x

\sqrt[3]{0.95} \approx 0.9833

\sqrt[3]{1.1} \approx 1.0333

Step-by-step explanation:

Given the function: g(x)=\sqrt[3]{1+x}

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial,L(x)=f(a)+f'(a)(x-a)

a=0

g(0)=\sqrt[3]{1+0}=1

g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}

Therefore:

L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x

(b)\sqrt[3]{0.95}= \sqrt[3]{1-0.05}

When x = - 0.05

L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833

\sqrt[3]{0.95} \approx 0.9833

(c)

(b)\sqrt[3]{1.1}= \sqrt[3]{1+0.1}

When x = 0.1

L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333

\sqrt[3]{1.1} \approx 1.0333

7 0
3 years ago
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