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lakkis [162]
3 years ago
12

HURRY PLEASE! i need this asap

Mathematics
1 answer:
denpristay [2]3 years ago
3 0

Answer:

4 \sqrt[3]{6x}^{2}

Step-by-step explanation:

a like radical just means one that's similar to the original. going through your options, you should look for one with the exact same contents in the brackets as the one asked in the question. in this case, we see it's the third one because the contents inside are the same. the outside value (or known as the stretch/compression value) doesn't change the fact that it is still like to the original.

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What type of function is represented by the table of values below?
GrogVix [38]

Answer:

C. Linear

Step-by-step explanation:

The function is going up by a constant rate of 4.

3 0
2 years ago
Read 2 more answers
What is this percent of change 40;72
salantis [7]
Given:
Original or base : 40
New amount : 72

We need to find the difference of the new amount and the original amount

72 - 40 = 32

We then divide the difference by the original amount

32/40 = 0.80

We multiply the quotient by 100%

0.80 x 100% = 80%

The percentage of change is 80%
6 0
3 years ago
What is the value of the x in this equation. 7x = 14 ​
S_A_V [24]

Answer:

2

Step-by-step explanation:

7x=14

\frac{7x}{7}  =  \frac{14}{7}

x=2

hope it's helpful ❤❤❤

THANK YOU.

7 0
3 years ago
Is 7 a factor of 14 yes or no ?
Alex73 [517]

Answer:

Yes

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
What is the answer please.... need help
mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
3 years ago
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