All categories

3 years ago

What is the domain of f(x) = 2|x − 1| + 3? A. {x | x ≥ 3} B. {x | x < 1} C. {x | x ≤ -1} D. {x | x = all real numbers}

Mathematics

1 answer:

arlik [135]3 years ago

8 0

Answer:

all real numbers

Step-by-step explanation:

f(x) = 2|x − 1| + 3

The domain is what values can we put in for x

There is no limit to what we can put in for x

therefore the domain is all real numbers

You might be interested in

Written as a simplified polynomial in standard form, what is the result when (3x+7)^2 is subtracted from 3?

nadezda [96]

Given:

$(3x+7)^2$ is subtracted from 3.

To find:

The result of the given statement in the standard form of the polynomial.

Solution:

We need to find the result when $(3x+7)^2$ is subtracted from 3.

So, the resulted polynomial can be written as:

$P(x)=3-(3x+7)^2$

On simplification, we get

$P(x)=3-((3x)^2+2(3x)(7)+(7)^2)$ $[\because (a+b)^2=a^2+2ab+b^2]$

$P(x)=3-(9x^2+42x+49)$

$P(x)=3-9x^2-42x-49$

$P(x)=-9x^2-42x-46$

Therefore, the resulted polynomial in standard form is $-9x^2-42x-46$ .

7 0

2 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

2 years ago

Given a line segment with endpoints A(16, 8) and B(1, 3) what are the coordinates of the

blondinia [14]

Answer:

The line segment partitioned two-fifths from A to B is (10,6)

Step-by-step explanation:

First point from A to B is (16,8)

than find the difference between A to B i.e B - A

(1,3)-(16,8) = (-15,-5)

To measure the (2/5) difference we will multiply (-15,-5) with (2/5) which is equal to (-6,-2)

Now Add the difference to the first coordinate (point A) gives

Point of division = (16,8)+(-6,-2)

Point of division = (16-6, 8-2)

Point of division = (10,6)

4 0

3 years ago

What is the median of this list of numbers? 6, 2, 7, 2, 3

Vika [28.1K]

The median means the number that is in the middle of the spnumber sequence in least to greatest. 1) number the numbers up. 2,2,3,6,7 2) start by crossing off the last from the greatest. Then you can see that 3 is the median.

4 0

3 years ago

Read 2 more answers

Find the slope of the line that passes through (50, -70) and (77, 62). Simplify your answer and write it as a proper fraction, i

-Dominant- [34]

Answer:

44/9 or 4.8888888888889

Step-by-step explanation: ¯\_(ツ)_/¯

4 0

3 years ago

Read 2 more answers