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mamaluj [8]
3 years ago
5

How many decimal places are in the product? 0. 053×2. 78.

Mathematics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer:

there are 4 decimal places

Step-by-step explanation:

you must solve the math

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Why is 6/8 greater than 5/8 but less than 7/8
My name is Ann [436]
Simple!

Since we can observe that they all have the same denominators (bottom part) we can just observe the numerator (top part)

6 is indeed greater than 5, yes? So we can safely say that

6/8 > 5/8

Now the for the second part. 

6 is smaller than 7, yes? So we can make the observation that

6/8 < 7/8

Hope that makes sense! (Be sure to thank me, plz)

P.S. If you want an example of unlike denominators, let me know. <span />
7 0
3 years ago
Is x and x2 like terms or unlike terms​
babymother [125]

Like terms" are terms whose variables (and their exponents such as the 2 in x2) are the same. In other words, terms that are "like" each other. Note: the coefficients (the numbers you multiply by, such as "5" in 5x) can be different.

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3 years ago
Can someone help me
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3 years ago
3.75 written as a word
tigry1 [53]
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7 0
3 years ago
Determine the number of real solutions for each of the given equations. Equation Number of Solutions y = -3x2 + x + 12 y = 2x2 -
rosijanka [135]

Answer:

Step-by-step explanation:

Our equations are

y = -3x^2 + x + 12\\y = 2x^2 - 6x + 5\\y = x^2 + 7x - 11\\y = -x^2 - 8x - 16\\

Let us understand the term Discriminant of a quadratic equation and its properties

Discriminant is denoted by  D and its formula is

D=b^2-4ac\\

Where

a= the coefficient of the x^{2}

b= the coefficient of x

c = constant term

Properties of D: If D

i) D=0 , One real root

ii) D>0 , Two real roots

iii) D<0 , no real root

Hence in the given quadratic equations , we will find the values of D Discriminant  and evaluate our answer accordingly .

Let us start with

y = -3x^2 + x + 12\\a=-3 , b =1 , c =12\\D=1^2-4*(-3)*(12)\\D=1+144\\D=145\\D>0 \\

Hence we have two real roots for this equation.

y = 2x^2 - 6x + 5\\

y = 2x^2 - 6x + 5\\a=2,b=-6,c=5\\D=(-6)^2-4*2*5\\D=36-40\\D=-4\\D

Hence we do not have any real root for this quadratic

y = x^2 + 7x - 11\\a=1,b=7,-11\\D=7^2-4*1*(-11)\\D=49+44\\D=93\\

Hence D>0 and thus we have two real roots for this equation.

y = -x^2 - 8x - 16\\a=-1,b=-8,c=-16\\D=(-8)^2-4*(-1)*(-16)\\D=64-64\\D=0\\

Hence we have one real root to this quadratic equation.

7 0
3 years ago
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