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3 years ago

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consider this quadratic equation x2+2x+7=21 the nuber of positive solutions to this equation are blank. the approximate value of

the greatest solution to the equation, rounded to the nearest hundredth, is blank.

1 answer:

S_A_V [24]3 years ago

6 0

Answer:

<em><u>x²</u></em><em><u>+</u></em><em><u>2</u></em><em><u>x</u></em><em><u>+</u></em><em><u>7</u></em><em><u>-</u></em><em><u>2</u></em><em><u>1</u></em><em><u>=</u></em><em><u>0</u></em>

<em><u> </u></em><em><u> </u></em><em><u>x²</u></em><em><u>+</u></em><em><u>2</u></em><em><u>x</u></em><em><u>-</u></em><em><u>1</u></em><em><u>4</u></em><em><u>. </u></em><em><u> </u></em><em><u>=</u></em><em><u>0</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>x²</u></em><em><u>+</u></em><em><u>2</u></em><em><u>x</u></em><em><u>. </u></em><em><u> </u></em><em><u>=</u></em><em><u>1</u></em><em><u>4</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>x²</u></em><em><u>+</u></em><em><u>2</u></em><em><u>x</u></em><em><u>+</u></em><em><u>1</u></em><em><u>. </u></em><em><u> </u></em><em><u>=</u></em><em><u>1</u></em><em><u>4</u></em><em><u>+</u></em><em><u>1</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>(</u></em><em><u>X+</u></em><em><u>1</u></em><em><u>)</u></em><em><u>²</u></em><em><u>. </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>=</u></em><em><u>1</u></em><em><u>5</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>X+</u></em><em><u>1</u></em><em><u>. </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>=</u></em><em><u>+</u></em><em><u>/</u></em><em><u>-</u></em><em><u>√</u></em><em><u>1</u></em><em><u>5</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>X=</u></em><em><u>-</u></em><em><u>1</u></em><em><u>+</u></em><em><u>/</u></em><em><u>-</u></em><em><u>√</u></em><em><u>1</u></em><em><u>5</u></em>

<em><u>X=</u></em><em><u>-</u></em><em><u>1</u></em><em><u>+</u></em><em><u>3.872983346207</u></em>

<em><u> </u></em><em><u> </u></em><em><u>=</u></em><em><u>2.872983346207</u></em><em><u>(</u></em><em><u>greate</u></em><em><u>st</u></em><em><u> </u></em><em><u>sol</u></em><em><u>ution</u></em><em><u>)</u></em><em><u> round</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>possible</u></em>

<em><u>X=</u></em><em><u>-</u></em><em><u>1</u></em><em><u>-</u></em><em><u>3.872983346207</u></em>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u>=</u></em><em><u>-</u></em><em><u>4.872983346207</u></em>

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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

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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;

$6\ years\ later = 500,000$

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$t = 6\ A = 500000$

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:

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

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

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2 years ago

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<em><u>Add 8 to both sides</u></em>

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X is greater than the value 1.64.

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