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3 years ago

Im confused with Direct Variation can some help me?

Mathematics

1 answer:

hichkok12 [17]3 years ago

3 0

Answer: i would say false bc they are just flipping the numbers at the top and at the bottom

but sry if im wrong

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Write the ratio as a fraction in simplest form, with whole numbers in the numerator and denominator.

maks197457 [2]

The answer is 20/28=5/7

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3 years ago

Two less than 1/4 the cube of x is y =

enyata [817]

Cube of x is x^3, according to question

Y= 1/4 x ^3 -2

8 0

3 years ago

What is x=15 a a function and please explain the process

Vika [28.1K]

<span>Simplifying 15 + 5x = 0 Solving 15 + 5x = 0 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-15' to each side of the equation. 15 + -15 + 5x = 0 + -15 Combine like terms: 15 + -15 = 0 0 + 5x = 0 + -15 5x = 0 + -15 Combine like terms: 0 + -15 = -15 5x = -15 Divide each side by '5'. x = -3 Simplifying x = -3</span>

5 0

3 years ago

PC-CSIL

pav-90 [236]

The equality used is Subtraction property of equality

Option A is correct

Step-by-step explanation:

We need to identify which option best explains or justifies Step 1.

Step 1 is: $-c = ax^2 + bx$

The given equation is: $0 = ax^2 + bx + c$

To get the equation for step 1 we need to subtract c on both sides of equation i.e using subtraction property of equality

$0-c = ax^2 + bx + c-c$

$-c = ax^2 + bx$

So, The equality used is Subtraction property of equality

Option A is correct.

Keywords: Solving Quadratic Equations

Learn more about Solving Quadratic Equations at:

#learnwithBrainly

5 0

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

6 0

2 years ago

Read 2 more answers