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2 years ago

State the domain and range of the following function:

Mathematics

1 answer:

mojhsa [17]2 years ago

7 0

Answer:

$Domain = \{-3,0,2,1,6,3\}$

$Range = \{4,6,-2,-3,-7,2\}$

Step-by-step explanation:

Given

$\{(- 3,4), (0,6), (2, - 2), (1, - 3), (6, - 7), (3, 2)\}$

Required

The domain and range

A function is represented as:

$Function = \{(x_1,y_1),(x_2,y_2),....(x_n,y_n)\}$

Where

$x \to$ domain

$y \to$ range

So, we have:

$Domain = \{-3,0,2,1,6,3\}$

$Range = \{4,6,-2,-3,-7,2\}$

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Solve this equation. Leave the answer as an improper fraction. 3x^2 + 2 = 11 - x^2

Nikolay [14]

Answer:

$x = \frac{3}{2}orx=-\frac{3}{2}$

Step-by-step explanation:

The given Equation is

$3x^{2} + 2 = 11-x^{2}$ .................(i)

Adding ( $x^{2} -11$ ) on both sides of (i)

$3x^{2}+2+x^{2}-11 = 11-x^{2}+x^{2}-11$

solving and cancelling out the terms will give

⇒ $4x^{2} -9=0$

⇒ $(2x)^{2} -(3)^{2} = 0$ ...................(ii)

Now we Know that

$(a)^{2} -(b)^{2} = (a-b)(a+b)$

Applying this on equation (ii)

$(2x-3)(2x+3)=0$

This will lead us to

Either 2x-3 =0 .......(iii) or 2x+3=0 ..........(iv)

Solving equation (iii) for value of x

2x - 3 =0

adding 3 on both sides

2x -3 +3 = 0+ 3

2x = 3

Cross multiplying gives

$x=\frac{3}{2}$

Solving equation (iv) for value of x

2x + 3 =0

adding -3 on both sides

2x -3 +3 = 0- 3

2x = -3

Cross multiplying gives

$x=\frac{-3}{2}$

so $x=\frac{3}{2}$ and $x=\frac{-3}{2}$

7 0

3 years ago

Explain why it is not reasonable to say that 4.23is less than 4.13

Alina [70]

It is not reasonable because the decimal placing on the 4.23 is greater then the one on 413. (2 is bigger then 1). When dealign with decimals it is important to remember that if the numbers before the decimal are the same for both numbers you need to look at the first decimal as though it's the first number. so we know 4 and 4 are the same. Next we look at the .23 and .13. We know that the number 2 is bigger then 1 so it's safe to say that 4.23 is larger then 4,13.

7 0

3 years ago

Read 2 more answers

Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)

zloy xaker [14]

Answer:

$k=0$

$\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}$

$\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}$

Step-by-step explanation:

We are given the function:

$\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}$

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

$\displaystyle (20) = 20e^{k(1)}$

Simplify:

$1= e^k$

Anything raised to zero (except for zero) is one. Therefore:

$k=0$

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

$\displaystyle (40) = 20e^{k(1)}$

Simplify:

$\displaystyle 2 = e^{k}$

We can take the natural log of both sides:

$\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}$

By definition, ln(e) = 1. Hence:

$\displaystyle k = \ln 2$

Therefore:

$2k+1 = 2\ln 2+ 1 \approx 2.3863$

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

$\displaystyle (10) = 20e^{k(1)}$

Simplfy:

$\displaystyle \frac{1}{2} = e^k$

Take the natural log of both sides:

$\displaystyle \ln \frac{1}{2} = k\ln e$

Therefore:

$\displaystyle k = \ln \frac{1}{2}$

Therefore:

$\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931$

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2 years ago

Which of the following is NOT a part of a DNA molecule?

Alex

Amino acids

Step-by-step explanation:

In a DNA molecule Your going to find what's called, Nucleotides, Nucleotides contain phosphate group, a sugar group, and a nitrogen base.

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3 years ago

Pls help me this is for my final text and I’m struggling really bad!!!

vodka [1.7K]

I’m not so sure I fully understand the question. But I would say the rate is 1.

I say this because the line on your graph is a y=mx+b equation.

For every unit it moves over on the X axis it moves the same unit on the Y axis.

I hope this helps.

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3 years ago