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3 years ago

Solve the quadratic equation 3x^2+x-5=0

Mathematics

1 answer:

irinina [24]3 years ago

5 0

-1 <span>± </span><span><span>√61 / 6. Negative 1, plus or minus square root of 61 divide by 6</span></span>

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3. There are 6 places set at each table for the big outdoor party.

Ainat [17]

You do 6x4 , The amount of legs on the chairs is irrelevant information, but your answer will be 24 people
24 people

8 0

3 years ago

Lilli jogs 12 mile in 110 hour What is Lilli's rate in miles per hour?

Verizon [17]

The problem requires us to find the rate or speed of Lilli in miles per hour. Therefore, we just have to divide the given data.

12/110 = 0.10909.... Therefore, Lilli's speed or rate is 0.11 mi/hr.

5 0

3 years ago

What is the unit rate of the proprrional relationship represented by the equation y=3x

NeX [460]

The unit rate of the proportional relationship y = 3x is 3

<h3>How to determine the unit rate?</h3>

The proportional relationship is given as:

y = 3x

Divide both sides by x

y/x = 3

For a proportional relationship y/x = k;

The unit rate is 3

This means that the unit rate of the proportional relationship y = 3x is 3

Read more about proportional relationship at

brainly.com/question/12242745

#SPJ1

6 0

2 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

7. Paula reduced the size of a rectangle to a height of 2 inches.

drek231 [11]

Answer: if the width was not effected then the width is still 24 inches

if the area stayed the same, then

find area

area=legnth times width

288=12 times 24

so the new height./legnth =2

288=2 times width

divide by 2

144=width'

the new width =144

6 0

3 years ago