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4 years ago

How do I solve this literal equation

Mathematics

1 answer:

dybincka [34]4 years ago

6 0

Multiply both sides by a
az=ab+m
Subtract ab from both sides
az-ab=m
pull a from az-ab
a(z-b)=m
Divide both sides by z-b
a=m/z-b

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A bag contains 36 marbles, some

vesna_86 [32]

Answer:

Step-by-step explanation:

Given ratio: 1:5

Total number of marbles = 36

Consider, number of Transparent marbles = 1x

number of green marbles = 5x

So,

1x + 5x = 36

6x = 36

x = 36/6

x = 6

So, the number of transparent marbles is:

1x = 1 x 6 = 6 transparent marbles.

7 0

3 years ago

Read 2 more answers

A bacteria colony increases in size at a rate of 4.0581e1.6t bacteria per hour. If the initial population is 36 bacteria, find t

Harrizon [31]

Answer:

1560

Step-by-step explanation:

The rate of Increase of the population (P) of the bacteria is given as:

$\frac{dP}{dt} =4.058e^{1.6t}$

$dP =4.058e^{1.6t}}dt\\Taking\: Integrals\\\int dP =4.058 \int e^{1.6t}dt\\P(t)=\frac{4.058}{1.6} (e^{1.6t} +K)\\P(t)=2.53625 (e^{1.6t} +K)$

Where k is a constant of Integration.

At t=0, P(t)=36

$36=2.53625 (e^{1.6*0} +K)\\36=2.53625 (e^{0} +K)\\36=2.53625 (1 +K)\\36=2.53625 +2.53625K\\K=13.19$

Therefore:

$P(t)=2.53625 (e^{1.6t} +13.19)\\At \: t=4\\P(4)=2.53625 (e^{1.6*4} +13.19)\\=1559.88\\P(4)=1560$

3 0

3 years ago

Help Solve and graph |-8x|<8<br> which is it

seraphim [82]

Answer:

The correct option is B.

Step-by-step explanation:

We know, from the definition of |x|,

|-8x| = - 8x, when - 8x ≥ 0 i.e. x ≤ 0 ......... (1)

and |-8x| = 8x, when - 8x < 0 i.e. x > 0 .......... (2)

Now, for x ≤ 0 {i.e. condition (1)}

- 8x ≤ 8, ⇒ x ≥ - 1 ......... (3)

Again, for x > 0 { i.e. condition (2)}

8x ≤ 8, ⇒ x ≤ 1 .......... (4)

So, combining conditions (3) and (4) we get -1 ≤ x ≤ 1.

So, the correct option is B. (Answer)

3 0

4 years ago

To add or subtract linear expressions, combine _______ terms. What is the answer to the first example?

just olya [345]

The answer is “like” terms.

You add common terms together to get variables by themselves.

6 0

3 years ago

What condition renders this true, where S and T are sets?

daser333 [38]

Answer:

There are no sets S, T for which

$S \cup T^c=S^c\cap T$

holds

Step-by-step explanation:

Let the set A be

$A=S\cup T^c$

By de De Morgan's Law

$A^c=S^c\cap ((T)^c)^c$

But

$((T)^c)^c=T$

$A^c=S^c\cap T$

We conclude that

$S \cup T^c=S^c\cap T\Rightarrow A=A^c$

which is a contradiction because no set is equal to its complement.

3 0

3 years ago