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3 years ago

Determine the ordered pair that satisfies the equation, -5x + 3y = 7.

Mathematics

1 answer:

sammy [17]3 years ago

3 0

Answer:

C) (7 , 14)

Step-by-step explanation:

Substitute each ordered pair into the equation

-5x + 3y = 7.

A) (2 , -1)

-5(2) +3(1) = -10 +3 = -7 this does not equal 7 so not true

B) (-5 , 3)

-5(-5) +3(3) = 25 +9 = 34 this does not equal 7 so not true

C) (7 , 14)

-5(7) + 3(14) = -35 + 42 = 7 this equals 7 so true

D) (-2 ,−135)

-5(-2) + 3(-135) =10+405 =415 this does not equal 7 so not true

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Marissa is painting one of her bedroom walls. She marks off 1/4 of the wall. Then she divides the marked section into 3 equal pa

Sergio039 [100]

Answer:

1/12

Step-by-step explanation:

The question can be written as (1/4) / 3. This is equivalent to (1/4)*(1/3), which is 1/12.

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3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

loris [4]

Answer:

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Step-by-step explanation:

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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

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2 years ago

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What is the solution of the system of equations? y = –4x + 10 y = –2x – 6

Elan Coil [88]

D. (8, -22) is the correct answer

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3 years ago

An oblique cylinder has a radius of 10 meters and a volume of 2,500 cubic meters. Use Cavalieri’s Principle to solve for the hei

shusha [124]

The idea of the Cavalieri's Principle states that even if the cylinder is oblique, so long as the planes in the figure are parallel, the equation would still be equal to,
V = Bh = πr²h
Substituting,
2,500 = π(10 m)²h
h = 7.95 m
The answer should be 5 meters.

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2 years ago