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3 years ago

Urgant !! 3(2x-4) - 6(x-3) please provide explanation

2 answers:

6 0

3 (2x-4)-6 (x-3)
Distribute by multiplication 3 to both 2x and -4
And distribute by multiplication -6 to both x and -3

6x-12-6x+18
Add like terms

6

DIA [1.3K]3 years ago

3 0

3(2x - 4) - 6(x - 3)
Distribute the 3 and the -6
6x - 12 -6x + 18
Combining like terms the expression simplifies to 6

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What is the simplified form of 20 times x to the sixth power over 15 times y to the fifth power times the fraction 5 times y squ

Delvig [45]

In order to easily picture out the operation involving fractions, let's write it in an algebraic equation. So, the above description is written as

20x⁶/15y⁵ × 5y²/6x⁴

The first thing to do is to simplify the easiest terms: the constants. Just multiply all the constants on the numerator and all the constants on the denominator independently. The simplified constant becomes

(20*5)/(15*6) = 10/9

So, the equation to be solved now is

10x⁶y²/9y⁵ x⁴

So, I also multiplied all the variables in the numerator and denominator independently, just like the constants. Then, you apply the laws of exponents. When you divide terms with exponents of the same value of the base, you subtract the exponent on the numerator minus the denominator. It goes like this:

10/9 * x⁶⁻⁴*y²⁻⁵ = 10/9*x²y⁻³

Now, when the exponent as written to the power of a negative exponent, just simply take its reciprocal to make it positive. So, the final answer is

10x²/9y³ or that is letter D.

8 0

4 years ago

Which one of these numbers is a perfect square? 22 15.6 16, 27<br> 15<br> 16

Simora [160]

16
square root of 16 is 4
4x4=16

5 0

4 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Penelope’s Printing Press promptly sails 5-Cent pencils and 10-cent pens to the public. It sold 1000 writing utensils and took i

Bumek [7]

Answer:

Number of pens and pencils sold are 490 and 510 respectively.

Step-by-step explanation:

Let, number of pens = x and the number of pencils = y.

It is given that,

Cost of one pen = 10 cents = 0.1 dollar

Cost of one pencil = 5 cents = 0.05 dollar

As, the total number of writing utensils sold are 100. So, we have,

$x+y=1000$

Also, the total sales is $74.50 gives,

$0.1x+0.05y=74.50$

Thus, the system of equations is,

x + y = 1000 .................................... (1)

0.1x + 0.05y = 74.50 .....................(2)

Multiply (1) by 0.1 and subtract both the equations, we get,

0.05y = 25.5 i.e. y= 510

So, x= 1000- 510 i.e. x= 490.

So, the number of pens and pencils sold are 490 and 510 respectively.

4 0

4 years ago

Suppose KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. Prove that A is equidistant from LM and KN.

Neporo4naja [7]

Answer:

This is proved by ASA congruent rule.

Step-by-step explanation:

Given KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. we have to prove that A is equidistant from LM and KN i.e we have to prove that AP=AQ

we know that the diagonals of parallelogram bisect each other therefore the the bisectors of ∠K and ∠L must be the diagonals.

In ΔAPN and ΔAQL

∠PNA=∠ALQ (∵alternate angles)

AN=AL (∵diagonals of parallelogram bisect each other)

∠PAN=∠LAQ (∵vertically opposite angles)

∴ By ASA rule ΔAPN ≅ ΔAQL

Hence, by CPCT i.e Corresponding parts of congruent triangles PA=AQ

Hence, A is equidistant from LM and KN.

7 0

3 years ago