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3 years ago

X^2-7x-6=0 Use the quadratic formula to complete the equation show step by step

Mathematics

2 answers:

vladimir1956 [14]3 years ago

7 0

x=(-b +- √b^2-4ac)/2a

x=7+-√49--24/2a

asambeis [7]3 years ago

3 0

Answer:

x = 7 ± √73 / 2

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Find the inverse of the function:<br> y=log₂(-4x)

Vlad1618 [11]

Answer:

-2^(x-2)

Step-by-step explanation:

Using differential calculus:

We know that the integral of 1/x = ln(x)

or 2x/ln(2) = log2(x), so we plug in our knowledge and chain rule to figure out that the answer to the inverse is -2^(x-2) :)

5 0

4 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0

2 years ago

Read 2 more answers

Need help asap!!!!!!!!!!!

salantis [7]

Answer:

the answer should be d for tour answer

3 0

3 years ago

Please please please please please please please help

Ilia_Sergeevich [38]

Answer:

a. 90 pack

b. $0.27

c. $0.18

d. $0.21

e. 60 pack

Step-by-step explanation:

a. i think the 90 pack container should be having least unit rate per pack,as purchasing in bulk amount the cost of a unit should be reduced to some amount which is must to lure the customer to purchase in bulk amounts.

b. unit price = total cost of container ÷ number of individual units inside it

for 20 pack, unit rate = 5.49÷20 = $0.27

c. for 60 pack, unit rate = 10.97÷60 =$0.18

d. for 90 pack, unit rate = 18.95÷90 =$0.21

e. ∴ the 60 pack container has the least unit rate per pack which is in contrary with our expectation of 90 pack to be lowest expecting bulk order.

5 0

3 years ago

How is the graph of y = |x| − 1 different from the parent function f(x) = |x|?

Jlenok [28]

Answer:

Step-by-step explanation:

the graph moves down one unit

if you add/subtract outside the parent function the graph moves up/down

3 0

3 years ago