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1 year ago

Pq=6x+25 and qr=16-3x; find pr

Mathematics

2 answers:

Nitella [24]1 year ago

8 0

Answer:

PR = 3x + 41

Step-by-step explanation:

Given

PQ = 6x + 25

QR = 16 - 3x

Q is the common point on the line PR, thus dividing PR into PQ and QR.

Therefore, we can write,

PR = PQ + QR

PR = 6x + 25 + 16 - 3x

PR = 6x- 3x + 25 + 16

PR = 3x + 41

Therefore, PR = 3x + 41

Viktor [21]1 year ago

3 0

$Given:\\\overline{PQ}=6x+25\\\overline{QR}=16-3x\\\\Finding:\overline{PR}$

$If \; Q \; is \; a \; point \; in \; between \; the \; line \; segment \; \overline{PR}, \; \\ then \; the \; distance \; of \; line \; segment \; \overline{PR} \; \\ can \; be \; found \; by \; adding \; line \; segment \; \overline{PQ} \; and \; \overline{QR}.$

$Using \; the \; concept \; \overline{PQ}+\overline{QR}=\overline{PR} \;,\\ we \; need \; to \; set \; up \; an \; equation \; given \; below:$

$\overline{PR} =6x+25 +16-3x\\\\Step \; 1: Grouping \; Like \; Terms\\\overline{PR} =6x-3x+25 +16\\\\Step \; 2: Combining \; Like \; Terms\\\overline{PR} =3x+41$

Conclusion:

$\overline{PR} \; is \; represented \; the \; expression \; 3x +41$

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Give an example of a qualitative variable and an example of a quantitative variable (discrete or continuous.) Explain the common

galben [10]

Answer:

Example of qualitative variable: hair colour.

Example of discrete quantitative variable: age.

a) Qualitative data displays are pie charts, histograms

b) Quantitative data displays are scatter and line graphs.

Step-by-step explanation:

A qualitative variable expresses a non-numerical quality of an object or person. For example, hair colour (brown, blonde, red...) or eye colour (green, blue, brown...).

A quantitative variable is a numerical value. For example, temperature (100 K, 2000 K...) or age (12 years, 20 years...).

A discrete quantitative variable can be obtained by counting, like the number of cars in a road. This is plotted in scatter graphs. For continuous variable, it can be obtained by measuring, like the height of your family members. This is plotted in line graphs.

Pie charts: is a circular graphic that shows the statistics or number of people or objects with certain characteristics. For example, how many people have brown hair, how many are blonde and how many are redheaded.
Histograms: they show vertical bars associated with the qualitative variable in the x-axis and the number of objects or people with that characteristic in the y-axis.
Scatter: it is a graph with x and y axis and using Cartesian coordinates. Since it is for quatities, numbers can be represented as points.
Line graphs: it is basically the same as a scatter plot but in this case the points can be joined by a line because the quantities are connected or are continuous.

6 0

1 year ago

Read 2 more answers

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 0

1 year ago

Find where the slope of the curve is defined:<br> x^2y-xy^2=4

Natali [406]

You do the implcit differentation, then solve for y' and check where this is defined.
In your case: Differentiate implicitly: 2xy + x²y' - y² - x*2yy' = 0
Solve for y': y'(x²-2xy) +2xy - y² = 0
y' = (2xy-y²) / (x²-2xy)
Check where defined: y' is not defined if the denominator becomes zero, i.e.
x² - 2xy = 0 x(x - 2y) = 0
This has formal solutions x=0 and y=x/2. Now we check whether these values are possible for the initially given definition of y:
0^2*y - 0*y^2 =? 4 0 =? 4
This is impossible, hence the function is not defined for 0, and we can disregard this.
x^2*(x/2) - x(x/2)^2 =? 4 x^3/2 - x^3/4 = 4 x^3/4 = 4 x^3=16 x^3 = 16 x = cubicroot(16)
This is a possible value for y, so we have a point where y is defined, but not y'.
The solution to all of it is hence D - { cubicroot(16) }, where D is the domain of y (which nobody has asked for in this example :-).
(Actually, the check whether 0 is in D is superfluous: If you write as solution D - { 0, cubicroot(16) }, this is also correct - only it so happens that 0 is not in D, so the set difference cannot take it out of there ...).
If someone asks for that D, you have to solve the definition for y and find that domain - I don't know of any [general] way to find the domain without solving for the explicit function).

3 0

2 years ago

What is the value of the product (4 – 2i)(6 + 2i)? 20 – 4i 20 + 16i 28 + 16i 28 – 4i

Darya [45]

(4 – 2i)(6 + 2i)

FOIL

First : 4*6 =24

Outer: 4*2i = 8i

Inner: -2i* 6 = -12i

Last: -2i*2i = -4i^2 =-4(-1) =4

Add together

24+8i-12i+4

28-4i

Answer: 28-4i

8 0

2 years ago

Read 2 more answers

Chandra created a budget matrix based on her regular and expected expenses for the year. Expense Jan. Feb. Mar. Apr. May June Ju

Ivan

Answer:

Chandra's Average Monthly Expenses are;

1) For Both Jan and Feb are $269

2) For the remaining 10 months each are $244

Step-by-step explanation:

From Chandra's matrix all monthly expenses are all same that is,

Cell phone $71, Rent $1,025, Gym $75, Internet $25, Auto insurance $425, Gas $ 120, Food $145. which are all the expenses carried out every month for 12 months.

That means Chandra carries out a total of 7 expenses for the month of January and February while she carried a total of 6 expenses for the remaining 10 month which was noted from the matrix that Auto insurance was carried out only in the month of January and February.

Therefore, you start by adding up each month total expenses, which are ;

January = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886

February = $71 + $1,025 + $75 + $25 + $425 + $120 + $145 = $1886

March = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

April = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

May = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

June = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

July= $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

August = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

September = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

October = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461

November = $71 + $1,025 + $75 + $25 + $120 + $145 = $1461