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2 years ago

7

4x-(12 divided 3) +2x=20

1 answer:

Savatey [412]2 years ago

3 0

Answer:

4

Step-by-step explanation:

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Solve the following equation. Then place the correct number in the box provided. 3 x = 12

Scorpion4ik [409]

Answer:

I believe 4 but I'm not that good at math but 4

3 0

2 years ago

Can someone help me with these 2 questions please and thank you?

Inessa05 [86]

2 × 3 = 6. 2. 7×3 =21

4 0

3 years ago

Solve for x<br> Urgent need help

fiasKO [112]

I believe you need to solve this using the quadratic formula!
To begin, this is what it is:
x= -b ± <span>√ b^2 - 4ac / 2a
Just plug in what you have in your problem...
2 being a, 13 being b, and -24 being c.
So we get:
x= -13 </span>± <span>√13^2 - 4(2)(-24) / 2(2)
x= -13 </span><span>± √169 - 8 (-24) / 4</span>
<span>x= -13 <span>± √169 + 192 / 4</span>
x= -13 </span>± √<span>361 / 4
The square root of 361 is 19.
So you have: -13 </span><span>± 19 / 4.
Here's where you take the equation </span>-13 <span>± 19 and put the addition and subtraction sign to use.
-13 - 19 = -32
and
-13 + 19 = 6
Now all is left to do is divide the two numbers by 4.
-32/4 = -8
and
6/4 = 3/2

x = -8, 3/2</span>

6 0

3 years ago

Help me..............

Virty [35]

The correct answer would be (C.) x>90

3 0

3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago