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2 years ago

Solve the equation using the quadratic formula -x^2+27=6x

Mathematics

2 answers:

san4es73 [151]2 years ago

8 0

$-x^{2}+27=6x\\-x^{2}-6x+27=0$

Quadratic formula

$\frac{-(-1)±\sqrt{6^{2}-4(-1)(27)}}{2(1)}$

$\frac{1±\sqrt{36+108}}{2}$

$\frac{1±\sqrt{144}}{2}$

$\frac{1±12}{2}$

$\frac{1+12}{2}=\frac{13}{2}=6.5\\\\\frac{1-12}{2}=\frac{11}{2}=-5.5$

6.5 or -5.5

MrMuchimi2 years ago

7 0

The answer is:

X= -9, 3

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Sarah made $180 for 12 hours of work. At the same rate, how much would he make for 17 hours of work?

lisabon 2012 [21]

Answer:

1260

Step-by-step explanation:

180 x 2

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2 years ago

What is RS? Please help will give Brainlyest.

IceJOKER [234]

Answer:

Step-by-step explanation:

RV=VU and SW=WT ⇒ $VW=\dfrac{RS+UT}{2}$

Therefore:

$3x+5=\dfrac{2x+15+6x-37}{2}\\\\3x+5=\dfrac{8x-22}{2}$

3x + 5 = 4x - 11 {subtract 5 from both sides)

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So:

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3 0

2 years ago

What is the value of 5(a + b) when a=1 and b=6

lana [24]

Answer: 35

Step-by-step explanation:

We substitute a for 1 and b for 6.

5(1+6)=5(7)=35

5 0

2 years ago

Please ' help me I need it please I'm begging you

vazorg [7]

Answer: 0.75 (choice B)

=============================================

Explanation:

One way to set up the equation is to think of it like this

vertical/horizontal = vertical/horizontal

So we could say

(3.5)/(28) = x/6

Cross multiply and solve for x

3.5*6 = 28*x

21 = 28x

x = 21/28

x = (3*7)/(4*7)

x = 3/4 in fraction form

x = 0.75 in decimal form

-------------

Another possible set up equation is

28/6 = (3.5)/x

in this case I divided the horizontal sides together (28 and 6) and the vertical sides divide to form their own separate fraction as well.

Solving that equation should lead you to x = 3/4 = 0.75

Other equations are possible.

8 0

1 year 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

1 year ago