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3 years ago

Why must ice cream be cooled below 0 before is freezes?

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

4 0

It could melt if it’s anything higher

djverab [1.8K]3 years ago

3 0

Zero Celsius is freezing temperature. It would melt if it were higher than zero.

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Y= -3x + 2 y = 2-3x What is the solution

coldgirl [10]

Answer:

the first question y=-6

the second question

y=1/2

Step-by-step explanation:

y=(3x2)

y=-6

the second question x+3x=2

4x=2

y=1/2

6 0

2 years ago

Baxter has a sandbox that holds Kinetic sand. The diagram

Ede4ka [16]

For any polygon:-

Volume=Area of base×Height

Here

Base is triangle

Area=

1/2(4)(3)
2(3)
6cm²

Volume:-

6×5
30cm²

6 0

2 years ago

The equation of the piecewise defined function f(x) is below. What is the value of f(1)?

cestrela7 [59]

Answer:

-1

Step-by-step explanation:

First, we need to find which expression to use for f(1). The first function does not include 1, because it is <, and not $\leq$

The second functions includes 1, because it has a $\leq$ , meaning greater than or equal to, so this is the function we must use.

-x^2

Plug 1 in for x

-(1^2)

Solve the exponent first

-(1)

-1

So, f(1)=-1

3 0

3 years ago

Read 2 more answers

Geometry math question no Guessing and Please show work thank you

nataly862011 [7]

If two lines are parallel and a transversal passes through them , then alternate interior angles are equal.

So here we have

3x+15=5x-5

5x-3x=15+5

2x=20

x=20/2=10

so option A . 10 is correct

6 0

3 years ago

Find from first principles the derivative of cos x

Lisa [10]

Answer:

Please see the explanation.

Step-by-step explanation:

Let

$f\left(x\right)=cosx$

By the first principle

$f\:'\left(x\right)=\lim _{h\to 0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$

$=\lim _{h\to 0}\left(\frac{cos\:\left(x+h\right)-cos\:x}{h}\right)$

$=\lim _{h\to 0}\left[\frac{cos\:x\:cos\:h-sin\:x\:sin\:h\:-\:cos\:x}{h}\right]$

$=\lim _{h\to 0}\left[\frac{-cos\:x\left(1-cos\:h\right)-sin\:x\:sin\:h\:}{h}\right]$

$=\lim _{h\to 0}\left[\frac{-cos\:x\left(1-cos\:h\right)\:}{h}-\frac{sin\:x\:sin\:h}{h}\right]$

$=-cosx\:\left(\lim \:_{h\to \:0\:}\frac{1-cos\:h}{h}\right)-sin\:x\:\lim \:\:_{h\to \:\:0}\:\left(\frac{sin\:h}{h}\right)$

$=-cosx\:\left(0\right)-sinx\left(1\right)$

$=-sin\:x$

8 0

3 years ago