in parallelogram pqsr what is pq

one of the rules for a parallelogram is that opposite sides are congruent.
so…all you need to do is set up the equation which is 2x + 5 = 4x + 1
move 2x to other side (meaning subtract 2x from both sides) to get 5 = 2x + 1
move 1 to other side to get 4 = 2x
divide both sides by 2 to get 2 = x
now all you need to do is plug it in the equation for PQ which is 2(2) + 5 to get 9

The value of side PQ of the given parallelogram is; PQ = 9

Parallelogram Properties

In parallelograms, we know that they usually have 4 sides. However 2 of the parallel sides are equal.

Thus, from the given parallelogram PQSR, we can say that;

PQ = RS

Thus;

2x + 5 = 4x + 1

Rearranging gives us;

4x – 2x = 5 – 1

2x = 4

x = 4/2

x = 2

Thus; PQ = (2x + 5) = 2(2) + 5 = 9

which are the solutions of x2 = –7x – 8?

You can start by rewriting the equation so that the right side equals zero. Add 

 and 

 to both sides.

You can now use the quadratic equation (below), where 

 and 

, to find solutions. Plug in these values for 

 and 

 into the equation and simplify.

The final answer is the combination of both solutions.

Or approximately…

x^2 = -7x -8

x^2 +7x +8 = 0

D = 49 -32 = 17

x_1,2 = (-7+/-sqrt17)/2 = (-7-sqrt17)/2 and (-7+sqrt17)/2

hope this will help you 

which missing item would complete this alpha decay reaction?

From the question,

Fermium which atomic mass of 257 decay through alpha to 253 Cf

The reaction is stated below.

257Fm –> 253Cf…+ 4He

100 98 2

Therefore,

253Cf —> 253Es + e-

98 . 99

What is Alpha decay?

Alpha decay is a form of radioactive decay that occur in the nucleus of an atom and the atomic nucleus emits an alpha particle which them decays into another atomic nucleus, with different mass number that is reduced by four.

Therefore, the completed part is

253Cf —> 253Es + e-

98 . 99

Learn more about alpha decay below.

This is alpha decay not beta decay. 

Fermium 257 decays by ALPHA decay into 253 Cf which has an atomic number of 98 

So your reaction is 

257Fm –> 253Cf…+ 4He 
100…………….98………2 

Interestingly Californium does undergo beta decay. 

253Cf —> 253Es + e- 
98……………99 

Hope this helps.

determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.

The empirical formula for a compound that is 36.86% N and 63.14% O by mass is N2O3.

HOW TO CALCULATE EMPIRICAL FORMULA?

The empirical formula of a compound can be calculated as follows:

  • N = 36.86% = 36.86g
  • O = 63.14% = 63.14g

First, we divide each value by its molar mass as follows:

  • N = 36.86g ÷ 14g/mol = 2.63mol
  • O = 63.14g ÷ 16g/mol = 3.95mol

Next, we divide by the smallest mole value;

  • N = 2.63mol = 2.63 = 1
  • O = 3.95mol ÷ 2.63 = 1.5

We multiply each value by 2 to get an approximate whole number ratio i.e. N = 2, O = 3

Therefore, the empirical formula of for a compound that is 36.86% N and 63.14% O by mass is N2O3.

Learn more about empirical formula at:

Mass of nitrogen = 14.0067
Mass of oxygen = 15.9994
In this compound nitrogen = 36.86 / 14.0067 = 2.63 
And oxygen = 63.14 / 15.9994 = 3.95 
now we have: N—– 2.63 and O—-3.95 
by dividing both with the smallest number we get
N——-2.63/2.63 = 1
O——-3.95/2.63 = 1.5
To get whole numbers we multiply both by 2
N= 1 x 2 = 2 And O = 1.5 x 2= 3
So, the empirical formula is N₂O₃.

installing addons and toolbars can slow down your browser’s performance.

This is absolutely true! Having more add-ons and toolbars means that your processor is going to be working much harder to function it all, while also trying to handle everything else it is getting. If you have to many add-ons/toolbars, it is going to be a constant fight on the IRQ and with a lower-end PC, your RAM may even fill up and crash your computer.

To put it simple, only add the toolbars and add-ons you need, and don’t add several extras.

the study and analysis of light according to its component wavelengths is called

ANSWER:

The study and analysis of light according to its component wavelengths is called spectroscopy.

EXPLANATION:

Spectroscopy is the branch of science  that is concerned with the investigation and measurement of spectrum produced when matter interacts with or emits electromagnetic radiation.It helps us to identify atoms and molecules in the object.Spectroscopy is used to find out Dopplers effect (the red shift and blue shift),which tells how fast the object is comming towards earth or moving away from the earth.

50+50-25×0+2+2

The correct answer using BODMAS for the evaluation of the question given is 104

We need to implement the concept of BODMAS on other to solve :

50 + 50 – 25 X 0 +2 +2

We perform the multiplication operation first :

-25 × 0 = 0

50 + 50 = 100

2 + 2 = 4

Hence,

50 + 50 – 25 X 0 +2 +2 = 100 + 0 + 4 = 104

Hence, the correct answer is 104

Answer:

The correct answer is 104

Step-by-step explanation:

Using BODMAS  rule we can find the answer

Each letter in the BOMAS indicate,

B – Bracket

O – of

D – Division

M – Multiplication

A – Addition

S – Subtraction

To solve the problem

The given problem is,  50 + 50 – 25 X 0 +2 +2

Using BODMAS rule we can write,

50 + 50 – 25 X 0 +2 +2 =  50 + 50 – 0 +2 +2

 = 104

The correct answer is 104

in the diagram of circle o, what is the measure of abc

Given:

Given that a circle O with two tangents BA and BC.

The major arc AC is 234°

The minor arc AC is 126°

We need to determine the measure of ∠ABC

Measure of ∠ABC:

We know the property that, “if a tangent and a secant, two tangents or two secants intersect in the interior of the circle, then the measure of angle formed is one half the difference of the measures of the intercepted arcs.”

Hence, applying the above property, we have;

Substituting the values, we get;

Thus, the measure of ∠ABC is 54°

Hence, Option b is the correct answer.

Answer:

My answer got deleted even though i was right.. B on edge

Step-by-step explanation:

54 degrees, dont ask questions

for which of the following mixtures will ag2so4(s) precipitate?

Answer:

c. 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)

d. 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)

Explanation:

Dissociation of Ag₂SO₄(s)    ⇄   2Ag⁺     +      SO₄²⁻

Solubility Product ;    = 

It is known that the solubility product    of Ag₂SO₄ = 1.2 × 10⁻⁵

If ionic concentration  <   ; precipitation will not occur

If  ionic concentration  >  ; precipitation will occur

Now, let’s pick the option one after the other to determine if precipitation occurs or not.

a).  

150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq)

moles of Ag = 

= 0.001 moles

moles of  = 

= 0.015 moles

New total volume = 150.0 mL + 5.0 mL = 155 mL

= 0.155 L

Molar Concentration of Ag:

 = 

Molar Concentration of   

 = 

Ionic Concentration   =   

 × 

∴    <   ; therefore precipitation will not occur.

b)

150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq)

moles of Ag = 

= 0.0015 mole

moles of  = 

= 0.015 moles

Molar Concentration of Ag:

 = 

Molar Concentration of   

 = 

Ionic Concentration   =   

∴    <   ; therefore precipitation will not occur.

c)

150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)

moles of Ag = 

= 0.002 moles

moles of  = 

= 0.015 moles

Molar Concentration of Ag:

 = 

Molar Concentration of   

 = 

Ionic Concentration   =   

∴    >   ; therefore precipitation will occur.

d)

150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)

moles of Ag = 

= 0.0025 moles

moles of  = 

= 0.015 moles

Molar Concentration of Ag:

 = 

Molar Concentration of   

 = 

Ionic Concentration   =   

∴    >   ; therefore precipitation will occur.

The mixture of Ag2SO4(s) that will precipitate:

c). 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)

d). 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)

Precipitation

This is known that Ag2SO4(s) dissociateds when:

(s)    ⇄    +  

The solubility  of   × 

Also,

The precipitation will take place only when:

  >  

so,

by measuring    of each of the given mixtures. we find that mixtures having  of   and  of 

and

mixture with   of   and  of  of  would precipitate.

Thus, options c and d are the correct answers.

Learn more about “Precipitation” here:

which interval for the graphed function contains the local maximum

Answer:

Local maximum = [3, 4]

Step-by-step explanation:

Step 1: Plot the given points [–1, 0] [1, 2] [2, 3] [3, 4] and draw the graph.

Step 2: Observe the graph, it is a straight line. Which has local minimum and local maximum at the closed interval.

Here with I have attached the graph.

The graphed function at local maximum at [3, 4]

Thank you.

The local maximum value of the function is in the interval of  and local maximum value of the function is in the interval of 

Further explanation:

Given:
The intervals are  and 

Explanation:
The given intervals are  and 

Plot the given points on the graph to obtain the local maximum value and the local minimum value in the given intervals.

The maximum value of the graphed function is 4 at 3.
The minimum value of the function is -1 at 0.

The local maximum value of the function is in the interval of  and local maximum value of the function is in the interval of 

Kindly refer to the imaged attached.

Learn more:
Learn more about inverse of the functionbrainly.com/question/1632445.
Learn more about equation of circle brainly.com/question/1506955.
Learn more about range and domain of the function brainly.com/question/3412497

Answer details:
Grade: High School
Subject: Mathematics
Chapter: Linear inequality

Keywords: interval, graphed, function, contains, local maximum, maximum, [-1,0], graphed function.