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2 years ago

A classmate simplifies the expression in exercise 1. is the expression he wrote equivalent to the original expression? explain.

1 answer:

4 0

Excercise 1:

No, this is not simplified fully. The full answer would be −7dk+14d−21k−7, not 14d - 9 - 21k - 7dk + 2. So, it is not equivalent.

Excercise 4:
7dk (2 - 3 - 1) - 7

Just multiply 7dk into (2) (-3) and (-1)
you'd get:
−14dk+−7 after simplifying it fully.

p = parameter
w = width

w+2.75

p = 2w+2(w+2.75) = 4w+5.5 = 30

Answer to the last one:

w = 6.125

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Heidi solved the equation 3(x + 4) + 2 = 2 + 5(x – 4). Her steps are below: 3x + 12 + 2 = 2 + 5x – 20 3x + 14 = 5x – 18 14 = 2x

saveliy_v [14]

step 1: distributive property.

step 2: combine like terms.

step 3: subtraction property of equality.

step 4: addition property of equality.

step 5: division property of equality.

hope this helps.

3 0

2 years ago

Read 2 more answers

Un padre para estimular a su hijo para estudiar matemáticas le dice: por cada ejercicio que resuelvas bien te daré 7 € y por cad

Naya [18.7K]

Answer:

8 ejercicios

Step-by-step explanation:

De acuerdo a la información dada, puedes plantear las siguientes ecuaciones:

x+y= 45 (1)

7x-1y= 19 (2)

x= ejercicios que resolvió bien

y= ejercicios que resolvió mal

Primero, puedes despejar x en (1):

x= 45-y (3)

Segundo, debes reemplazar (3) en (2):

7(45-y)-1y=19

315-7y-1y= 19

315-8y= 19

315-19= 8y

296= 8y

y= 296/8

y= 37

Tercero, debes reemplazar el valor de y en (3) para encontrar x:

x= 45-37

x= 8

De acuerdo a esto, la respuesta es que el hijo ha resuelto bien 8 ejercicios.

8 0

3 years ago

What is -2(4g-3)=30 I can't figure it out so I need so help

timofeeve [1]

Answer:

$g = - 3$

Step-by-step explanation:

$- 2(4g - 3) = 30$

$- 8g + 6 = 30$

$- 8g + 6 - 6 = 30 - 6$

$- 8g = 24$

$\frac{ - 8g}{ - 8} = \frac{24}{ - 8}$

$g = - 3$

4 0

3 years ago

Read 2 more answers

The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am

posledela

Step-by-step walkthrough:

Well a standard half-life equation looks like this.

$N = N_0 * (\frac{1}{2})^{t/p$

$N_0$ is the starting amount of parent element.

$N$ is the end amount of parent element

$t$ is the time elapsed

$p$ is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days$

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2} = 37g$

Now we just gotta do some algebra. Use the original function but this time, replace $N(t)$ with 10g and solve algebraically.

$10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}$

Take the log of both sides.

$log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})$

Use the exponent rule for log laws that, $log(b^x) = x*log(b)$

$log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})$

$\frac{log(\frac{5}{37})}{log(\frac{1}{2})} = \frac{t}{2.8days}$

$2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})} = t$

slap that in your calculator and you get

$t = 8.1 days$

7 0

1 year ago

HELP!!! 20 POINTS!!! URGENT!!!

VashaNatasha [74]

We know that :

$\clubsuit$ $ln(A) - ln(B) = ln(\frac{A}{B})$

$\clubsuit$ $aln(x) = ln(x)^a$

$\clubsuit$ $ln(A) + ln(B) = ln(AB)$

Using above ideas we can solve the Problem :

⇒ $\frac{3}{8}ln(x + 3) = ln(x + 3)^\frac{3}{8}$

⇒ $ln(x - 3) - ln(x + 3)^\frac{3}{8} = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]$

⇒ $4ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}] = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]^4 = ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}]$

⇒ $\frac{1}{3}lnx + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln(x)^\frac{1}{3} + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln[\frac{\sqrt[3]{x}(x - 3)^4}{\sqrt{(x + 3)^{3}}}]$

Option 3 is the Answer

8 0

2 years ago