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2 years ago

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What is the answer to this

1 answer:

kvv77 [185]2 years ago

7 0

A is correct
$3x + (2x + 3) = 5x + 3$

good luck

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need help.. The midpoint of GH is M(-6,-3). One endpoint is H (-4,4) Find the coordinates of endpoint G.

iren2701 [21]

Answer:

The coordinates of G are: (-8, -10)

Step-by-step explanation:

Mid point formula:

$x = \dfrac{x_1+x_2}{2}\\y = \dfrac{y_1+y_2}{2}$

Where,

$(x,y)$ are the coordinates of the mid point of the points $(x_1,y_1)$ and $(x_2,y_2)$ .

As per question statement, we are given that:

$x_2=-4\\y_2=4$

and

$x=-6\\y=-3$

To find:

$x_1 = ?\\y_1 = ?$

Solution:

Let us use the mid point formula:

$-6 = \dfrac{x_1-4}{2}\\\Rightarrow -12=x_1-4\\\Rightarrow x_1=-12+4\\\Rightarrow \bold{x_1=-8}$

$-3 = \dfrac{y_1+4}{2}\\\Rightarrow -6=y_1+4\\\Rightarrow y_1=-6-4\\\Rightarrow \bold{y_1=-10}$

The coordinates of G are: (-8, -10)

5 0

2 years ago

Can 9/43 besiplified and can 34/43 be simplified?

diamong [38]

No be because the denominator it an odd number and you can't divide an odd number hope i helped in anyway
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6 0

2 years ago

I need help pls answer right away

Zielflug [23.3K]

Answer:

Step-by-step explanation:numerator divided by denominator???

6 0

2 years ago

Read 2 more answers

I like to give easy points so what is 35+50+20 super easy

BARSIC [14]

105 cuz it is........

4 0

2 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

6 0

1 year ago

Read 2 more answers