All categories

3 years ago

Please answer this before 5 hours have passed ^.^

Mathematics

1 answer:

Rashid [163]3 years ago

3 0

Answer:

24m + 12 - - - (1)

12(2m + 1) - - - (2)

Step-by-step explanation:

Given the expression :

4(m + 3 + 5m)

Equivalent expressions include :

24m + 12 - - - (1)

12(2m + 1) - - - (2)

Using algebaraic properties :

4(m + 3 + 5m)

We can expand :

4m + 12 + 20m

4m + 20m + 12

24m + 12

Hence,

4(m + 3 + 5m) = 24m + 12

Using Substitution:

Let m = 5

4(m + 3 + 5m)

4(5 + 3 + 5(5))

4(5 +3 + 25)

4(33)

= 132

12(2m + 1)

12(2(5) + 1)

12(10 + 1)

12(11)

= 132

You might be interested in

What is 12,430,000 in expanded form

kotykmax [81]

I'm not sure exactly which type you are meaning since there are four types of expansions, but depending on which specific type here are your answers! :)

7 0

3 years ago

Read 2 more answers

Help meee, I need to get this done bu today but I have no idea

Hunter-Best [27]

Function a is linear, the equation is y = 5/3x or y=5/3x+0

3 0

4 years ago

What is the value of the expression above when x = 3 and y = 4? You must show all work and calculations.

spin [16.1K]

3*3+1/4*4^2
3*3+1/4*16
9+1/64
9.01562 I think is the answer

6 0

3 years ago

-4(1.75+1x)=18 What does x equal

77julia77 [94]

Answer:

x = -6.25

Step-by-step explanation:

-4(1.75 + 1x) = 18

1.75 + 1x = 18/-4

1.75 - 1.75 + x = -4.5 - 1.75

x = -6.25

Answer: x = -6.25

8 0

4 years ago

The output of an economic system Q, subject to two inputs, such as labor L and capital K, is often modeled by the Cobb-Dougl

Anna35 [415]

Answer:

Step-by-step explanation:

Given

Economic system Q is given by

$Q=cL^aK^b$

also $a+b=1$

if $Q=12,200$

$a=\frac{1}{6}$

$b=\frac{5}{6}$

$c=42$

substitute these values

$12,200=42\times (L)^{\frac{1}{6}}K^{\frac{5}{6}}$

$(L)^{\frac{1}{6}}K^{\frac{5}{6}}=\frac{12,200}{42}$

$K^{\frac{5}{6}}=\frac{12,200}{42(L)^{\frac{1}{6}}}$

$K=(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^{5}}$

differentiate w.r.t to L to get $\frac{dK}{dL}$

$\frac{dK}{dL}=(\frac{12,200}{42})^{\frac{6}{5}}\times (-5)\times L^{-6}$

$\frac{dK}{dL}=-5(\frac{12,200}{42})^{\frac{6}{5}}\times \frac{1}{L^6}$

$\frac{dK}{dL}=-\frac{4515.466}{L^6}$

6 0

3 years ago