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k0ka [10]
2 years ago
14

What is the same as 1.5 in a fraction

Mathematics
2 answers:
jeka942 years ago
4 0

Answer:

the answers is 3 over 2 in fraction form

xeze [42]2 years ago
4 0

Answer:

1 1/2 or 3/2

Steps:

As we know .5 = 1/2

If we take one the 1 out of 1.5 we can use it to connect 1 and 1/2

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See if you guys can figure this out.
Aleonysh [2.5K]
There isn't anything there
6 0
2 years ago
phil ate three less than twice as many pancakes as sara. together they combined to eat 15 pancakes. How many pancakes did each e
Levart [38]

Answer:

Let x represents the Sara eat pancake and y represents the Phil eat the pancake.

As per the statement:

Phil ate three less than twice as many pancakes as Sara.

"Twice as many pancake as Sara" means 2x

"three less than twice as many pancakes as Sara" means 2x -3

⇒y = 2x-3         .....[1]

It is also given that together they combined to eat 15 pancakes.

⇒x+y =15      .....[2]

Substitute equation [1] into [2] we get;

x+2x-3 = 15

Combine like terms;

3x-3 =15

Add 3 to both sides we get;

3x = 18

Divide both sides by 3 we get;

x = 6

Substitute the value of x in [1] we get;

y = 2(6)-3 = 12-3 =9

y = 9

Therefore, the pancakes did each eat are;

Sara eat pancake = 6 and Phil eat pancake = 9


8 0
3 years ago
Plz answer this <br> 1.580 ml: 1.12 liter: 104ml
tatyana61 [14]

Answer:

??

Step-by-step explanation:

4 0
3 years ago
For what value of constant c is the function k(x) continuous at x = 0 if k =
nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0
2 years ago
26. -2, 3, 8, 13, 18,...
-Dominant- [34]

Answer:

each number is being added by 5

Step-by-step explanation:

5 0
3 years ago
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