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seropon [69]
4 years ago
8

How do you know six over six and one are equivalent

Mathematics
2 answers:
Afina-wow [57]4 years ago
7 0
They are equal because let's say you have 6 slices of pizza and you and 5 friends came over to your house to have pizza all of you had one slice because their were 6 slices in the beginning and you and your 5 friends had one slice so that is 1 whole pizza that you and you friend ate
Leto [7]4 years ago
7 0

6/6 = 1

If you divide 6 by 6 it equals 1 that's why the statement is true;

6/6 = 1

1 = 1

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(5x^2+3x+4)-(2x^2+5x-1)
DedPeter [7]
Answer:
The answer is (3x^2-2x+5)
6 0
4 years ago
Put in simplest form 8/15 × 5/6
algol13
8/15 x 5/6
•multiply across the numerator and denominator.

= 40/90
• then find a common factor that will go into both the numerator and denominator equally.

Common factor: 5

= 8/18
•there is another common factor, which is 2

Simplified:

=4/9
6 0
4 years ago
Help me do question 9 and 10 I am 6th grade. Please
Phoenix [80]

Answer:

Step-by-step explanation:

9 - ratio 3:5

10 1/8 + 1/4 = 1/8 + 2/8 = 3/8 of the pizza eaten

so 5/8th of the pizza is left

8 0
3 years ago
HOW DO I FIND QR<br><br> PLEASE HELP
Paul [167]

Solution: For finding QR we need to apply Pythagoras Theorem

What is Pythagoras Theorem ?

ans : Pythagoras Theorem is the sum of square of two sides which is equal to the third side or the hypotenuse. This formula is valid oy incase of Right-Angled Traingle because one of three angles here is 90°

According to this law let's apply it in the diagram shown here.

  • (Hypotenuse)² = (Adjacent)² + (Opposite)²

  • (PR)² = (QR)² + (PQ)²

  • (5x - 2)² = (QR)² + (3x - 1)²

  • (QR)² = (5x - 2)² - (3x - 1)²

After factorising both of them we get

  • (QR)² = 25x² - 20x + 4 - (9x² - 6x + 1)

  • (QR)² = 25x² - 20x + 4 - 9x² + 6x - 1

  • (QR)² = 16x² - 14x + 3

  • QR = √(16x² - 14x + 3)

So, QR is √(16x² - 14x + 3)

8 0
3 years ago
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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