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Ronch [10]
3 years ago
14

Factor completely. 3x^2+30x+75=

Mathematics
2 answers:
frutty [35]3 years ago
8 0

Answer:

3(x+5)^2

Step-by-step explanation:

Talja [164]3 years ago
3 0

Answer:

3(x+5)(x+5)

Step-by-step explanation:

Factor 3x^2+30x+75

3x^2+30x+75

=3(x+5)(x+5)

Hope this helps!

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Porter believed that two additional variables can influence the national diamond along with the four original attributes. What a
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Answer:

Chance and government

Step-by-step explanation:

Chance events, such as major innovations, can reshape industry structure and provide the opportunity for one nation's firms to replace another's

Government, by its choice of policies, can detract from or improve national advantage

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4 years ago
Jerry had $1,018.00 in savings. Last week he spent $84.00 on a desk and $372.00 on a computer from his savings. This week at a p
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The correct answer is$412.00
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3 years ago
Enter the value of P so that the expression 3 (4 + n) is equivalent to 3 (n+p).
Gala2k [10]

Answer:

3 (4+n)=0

3×4+3×n=0

12+3n=0.... equation 1

3 (n+p)

3×n+3×p=0

3n+3p=0........... equation 2

eq 1 multiply 3 eq 2 multiply 3

36+9n=0

9n+9p=0

_ _ =0

_9n is +9n Crush

36+9p=0

p=36÷9

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4 0
3 years ago
Use Newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 + x +
rjkz [21]

Answer:

The value of x_2=-1.6923.

Step-by-step explanation:

Consider the provided information.

The provided formula is f(x)=x^3+x+6

Substitute x_1=-2 in above equation.

f(x_1)=(-2)^3+(-2)+6

f(x_1)=-8-2+6

f(x_1)=-4

Differentiate the provided function and calculate the value of f'(x_1)

f'(x)=3x^2+1

f'(x)=3(-2)^2+1

f'(x)=13

The Newton iteration formula: x_2=x_1-\frac{f(x_1)}{f'(x_1)}

Substitute the respective values in the above formula.

x_2=-2-\frac{(-4)}{13}

x_2=-2+0.3077

x_2=-1.6923

Hence, the value of x_2=-1.6923.

6 0
3 years ago
For thousands of years, gold has been considered one of the Earth's most precious metals. One hundred percent pure gold is 24-ka
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The answer is on bitly im about to put the link
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