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4 years ago

K is the midpoint of JL,JL=4x-2 and jk=7,find x KL and JL

1 answer:

4 0

K is the midpoint of JL...so JK=KL=7
JL=(JK + KL)>>>>>>whole part axiom
or,4x=2×JK>>>>>>>{JK=KL➡JK+KL=JK+JK}
or,4x=2×7
or,x=14/4=7/2

[ x=7/2 ] [KL=JK=7] and [JL=4x=(4×7)/2=14]

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What is an expression to represent the equation

Tatiana [17]

Answer:

8 + $\frac{x}{4}$

Step-by-step explanation:

I'm assuming you want an answer to the top section of the question.

We can start with 8.

When a problem asks for the sum, it's looking for addition, so we use the + sign.

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The problem says "a number," so let's represent that with x. Basically, x just means a number we don't know.

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8 + $\frac{x}{4}$

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3 years ago

Solve for equation x. <br> 3In(x)+2In(4)=In(128)

Serggg [28]

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3 0

3 years ago

The average discharge at the mouth of the Amazon River is 4,900,000 cubic feet per second. How much water is discharged from the

Misha Larkins [42]

Answer: Average discharge at the mouth of the Amazon River in 1 hour = $1.764\times 10^{10}$ cubic feet

Average discharge at the mouth of the Amazon River in 1 year= $3.66912\times10^{13}$ cubic feet[/tex]

Step-by-step explanation:

Given : The average discharge at the mouth of the Amazon River =4,900,000 cubic feet per second.

i.e. The average discharge at the mouth of the Amazon River in 1 second= 4,900,000 cubic feet

Since 1 hour = 3600 seconds

Then , the average discharge at the mouth of the Amazon River in 1 hour =

$4900000\times3600=17640000000$ cubic feet

Since , there are 2080 hours in a typical year.

Then, the average discharge at the mouth of the Amazon River in 1 year=

$2080\times17640000000\\\\=36,691,200,000,000=3.66912\times10^{13}$ cubic feet

6 0

3 years ago

When electricity (the flow of electrons) is passed through a solution, it causes an oxidation-reduction (redox) reaction to occu

oee [108]

Answer:

a. 135 g

b. 60.6 min

Step-by-step explanation:

a. What mass of Cu(s) is electroplated by running 28.5 A of current through a Cu2+ (aq) solution for 4.00 h? Express your answer to three significant figures and include the appropriate units.

The chemical equation for the reaction is given below

Cu²⁺(aq) + 2e⁻ → Cu(s)

We find the number of moles of Cu that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 28.5 A and t = time = 4.00 h = 4.00 × 60 min/h × 60 s/min = ‭14,400‬ s

So, n = It/F = 28.5 A × ‭14,400‬ s/96485 C/mol = ‭410,400‬ C/96485 C/mol = 4.254 mol

Since 2 moles of electrons deposits 1 mol of Cu, then 4.254 mol of electrons deposits 4.254 mol × 1 mol of Cu/2 mol = 2.127 mol of Cu

Now, number of moles of Cu = n' = m/M where m = mass of copper and M = molar mass of Cu = 63.546 g/mol

So, m = n'M

= 2.127 mol × 63.546 g/mol

= 135.15 g

≅ 135 g to 3 significant figures

b. How many minutes will it take to electroplate 37.1 g of gold by running 5.00 A of current through a solution of Au+(aq)?

The chemical equation for the reaction is given below

Au⁺(aq) + e⁻ → Au(s)

We need to find the number of moles of Au in 37.1 g

So, number of moles of Au = n = m/M where m = mass of gold = 37.1 g and M = molar mass of Au = 196.97 g/mol

So, n = m/M = 37.1 g/196.97 g/mol = 0.188 mol

Since 1 mol of Au is deposited by 1 moles of electrons, then 0.188 mol of Au deposits 0.188 mol of Au × 1 mol of electrons/1 mol of Au = 0.188 mol of electrons

We find the time it takes to deposit 0.188 mol of electrons that are deposited from

nF = It where n = number of moles of electrons, F = Faraday's constant = 96485 C/mol, I = current = 5.00 A and t = time

So, t = nF/It

= 0.188 mol × 96485 C/mol ÷ 5.00 A

= ‭18173.30‬ C/5.00 A

= 3634.66 s

= 3634.66 s × 1min/60 s

= 60.58 min

≅ 60.6 min to 3 significant figures

6 0

3 years ago

In triangle ABC, angle A = 2x°, angle B = (x-45)° and angle C = (x-15)°.

Lelechka [254]

$Sum\ of\ angles\ in\ triangle\ is\ equal\ to\ 180^{\circ}\\\\
To\ find\ x\ sum\ all\ angles:\\
2x+x-45+x-15=180\\\\
4x-60=180\ \ \ |Add\ 60\\\\
4x=60+180\\\\
4x=240\ \ \ |Divided\ by\ 4\\\\
x=60^{\circ}\\\\
\angle \ A=2x=2*60=120^{\circ}\\
\angle\ B=x-45=60-45=15^{\circ}\\
\angle\ C=x-15=60-15=45^{\circ}\\\\Triangle\ is\ \textbf{obtuse}.$

5 0

3 years ago