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3 years ago

1. Write fraction as a mixed number. 5/4

Mathematics

2 answers:

anygoal [31]3 years ago

7 0

1 1/4

4 fits into five once, leaving 1/4 left making it a mixed number.

Juli2301 [7.4K]3 years ago

6 0

1 1/4 do 5-4= 1 which means one is left over out of 4 so 1 1/4

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Geometry - Will give 5 stars + Thanks!

Fynjy0 [20]

Answer:

Step-by-step explanation:

We know:

The diagonals in a parallelogram divide by halves.

Therefore KG = UG and DG = CG.

We have

KG = 5y - 8, UG = 3y, DG = 4x - 7, CG = x + 2

Substitute:

5y - 8 = 3y add 8 to both sides

5y = 3y + 8 subtract 3y from both sides

2y = 8 divide both sides by 2

y = 4

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

4x - 7 = x + 2 add 7 to both sides

4x = x + 9 subtract x from both sides

3x = 9 divide both sides by 3

x = 3

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4 years ago

How many times does 7 go into 40

choli [55]

Answer:

5.7 times

Step-by-step explanation:

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4 years ago

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saveliy_v [14]

Answer:a and d

Step-by-step explanation:

7 0

3 years ago

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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

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3 years ago

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Find the value of p (x) = 4x^3 - 7x^2 - 5x + 43 if x= -3

Alla [95]

Answer:

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3 years ago

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