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Elis [28]
3 years ago
5

Will mark the brainliest

Mathematics
1 answer:
Sati [7]3 years ago
6 0

Answer:

B.   60x^8

Step-by-step explanation:

4x^2\sqrt{5x^4} \cdot 3\sqrt{5x^8} =

= 4x^2 \times 3 \sqrt{5x^4 \times 5x^8}

= 12x^2 \sqrt{25x^{4+8}}

= 12x^2 \sqrt{25} \sqrt{x^{12}}

= 12x^2 \times 5 \times x^6

= 60x^8

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2 (y+6) + 6 (y+6) Expand and simplify
san4es73 [151]
Hi there! The answer is 8y + 48.

2(y + 6) + 6(y + 6)
First work out the parenthesis.

2y + 12 + 6y + 36
Rewrite the expression in order to collect the terms.

2y + 6y + 12 + 36
Next up, collect the terms.

8y + 48
Now we've found our answer.
5 0
3 years ago
The given expression represents the area. Find the side length of the square.
Vesnalui [34]

9514 1404 393

Answer:

  (4x +5)

Step-by-step explanation:

The square of a binomial is ...

  (a +b)² = a² +2ab +b²

We assume that the given trinomial is a perfect square, so we can find the terms of the squared binomial by looking at the square roots of the first and last terms:

  √(16x²) = 4x

  √25 = 5

So, our first choice of side length for the square is ...

  s = (4x +5) . . . . . side length of the square

__

Using the above identity, we can verify that the square of this gives the right area:

  A = s² = (4x +5)² = (4x)² +2(4x)(5) +5²

  A = 16x² +40x +25 . . . . . . as required

8 0
3 years ago
What is 3t-2(t-1) is greater than or equal to 5t-4(2+t)?
Harman [31]
So the equation is    t-2 ≥ t-8 which also equals    t ≥ t-10 id ont know if this help but what are you trying to find
3 0
3 years ago
What is 0.32 written as a decimal
vredina [299]
Decimal: .32
fraction: 32/100 which, when reduced, is 8/25 
5 0
4 years ago
Read 2 more answers
Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
4 years ago
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