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mixer [17]
2 years ago
15

What is the LCD of 3/10 and 7/15??

Mathematics
2 answers:
Maurinko [17]2 years ago
8 0
Least Common Denominator of 10 and 15 is 30.So 30 might be the denominator.Hope this helped.
Nitella [24]2 years ago
3 0
30 would be a great answer

Hope it helps!.!.!.!.!.!.!
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FX) is defined by the equation f(x) = 4x2 - 2x +17. What effect will multiplying
iragen [17]

Step-by-step explanation:

the graph will be compressed vertically

6 0
3 years ago
Identify the function as either a constant, direct variation, absolute value or greatest integer function f (x)=-4
boyakko [2]
Answer: Constant

No matter what the input x is, the output f(x) is going to be -4. Therefore, the output is constant. 

Note: f(x) can be interchanged with y. So we can say y = -4.
7 0
3 years ago
please help me I only have 2 minutes left winner gets brainliest is 1 + (8r+9) and (2+8) + 8r equivalent
s344n2d4d5 [400]

Answer:

Yes, this is equivalent.

Hope this helps!

7 0
3 years ago
F(x) = 3x + x3
____ [38]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

\frac{d}{dx}\left(3x+x^3\right)

\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'

=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

solving

\frac{d}{dx}\left(3x\right)

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'

=3\frac{d}{dx}\left(x\right)

\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1

=3\cdot \:1

=3

now solving

\frac{d}{dx}\left(x^3\right)

\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}

=3x^{3-1}

=3x^2

Thus, the expression becomes

\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

                    =3+3x^2

Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value  f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 =  3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

Thus, the value of x​ will be:

x=2,\:x=-2

5 0
3 years ago
PLEASSSSSEEE HELPPP ME
Sonja [21]

I hope this helps you

3 0
2 years ago
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