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3 years ago

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Janet has 23 cups of potting soil for her new plants. she evenly distributes the soil into 5 pots. which location on a number li

ne correctly indicates the number of cups of potting soil that will be in each pot?
a.exactly 4
b.between 4 and 5
c.exactly 5
d.between 5 and 6

2 answers:

vodka [1.7K]3 years ago

7 0

The answer is between 4 and five because 23 divided by 5 is 4.6

larisa [96]3 years ago

4 0

We have to divide 23 by 5.
23/5=4.6
B. Between 4 and 5, is correct.

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3. Given ABCD ≅ FGHJ and mC = 9x - 7, mH = 5x + 13, find the value of x and the measures of angle C and angle H. Show all work

kondor19780726 [428]

Answers:

x = 5

m∠C = m∠H = 38 degrees

WORKINGS

Given that ABCD ≅ FGHJ

We know that corresponding angles of two congruent quadrilaterals are equal

Therefore,

m∠A = m∠F

m∠B = m∠G

m∠C = m∠H

m∠D = m∠I

Given,

m∠C = 9x – 7

m∠H = 5x + 13

Since m∠C = m∠H

9x – 7 = 5x + 13

Add 7 to both sides of the equation

9x – 7 + 7 = 5x + 13 + 7

9x = 5x + 20

Subtract 5x from both sides of the equation

9x – 5x = 5x – 5x + 20

4x = 20

Divide both sides of the equation by 4

4x/4 = 20/4

x = 5

To determine the measures of angle C and angle H

m∠C = m∠H

We know that m∠C = 9x – 7

Since x = 5

m∠C = 9(5) – 7

m∠C = 45 – 7

m∠C = 38

Therefore, m∠C = m∠H = 38 degrees

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3 years ago

Given that point j is the incenter of triangle nop, find jk if jm = x-2 and jl = 3x - 8

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3 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

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2 years ago

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