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1 year ago

I need help with 7 and 4 and 9 also 10 I will give the most higher points I got

Mathematics

1 answer:

Serga [27]1 year ago

8 0

7. 9/5 = 1 4/5
*When multiplying with a whole number u can turn it into a fraction by putting a one under it
So 3/1 times 3/5 = 9/5
*just multiply straight across

4. 2/1 times 1/4 = 2/4 = THEN SIMPLIFY = 1/2

9. 3/4 times 3/1 (cups)
*multiply straight across
9/4 = 2 1/4 (cups of sugar)
So she uses less than 3 cups to make all 3 loaves

* 10. Sorry I can’t tell if it says 5/6 on both of them
If it says 5/6 on both of them then the answer is obvious and she worked more than that over the course of 2 days

* 10. It must be a 5/9 on the second one
So if so
5/6 times 2/1 (days)
10/6 then we need to multiply the 6 denominator (bottom of the fraction) with something to = 9 denominator
9 divided by 6 = 1.5
10/6 times 1.5 for both the numerator (top) and denominator (bottom)
= 15/9 (simplified 1 6/9)
This one she exercised more than 5/9

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Answer:

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Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

$a\cdot x + b\cdot y + c\cdot z = d$

Where:

$x$ , $y$ , $z$ - Orthogonal inputs.

$a$ , $b$ , $c$ , $d$ - Plane constants.

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For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

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$y = m\cdot x + b$

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - x-Intercept, dimensionless.

If $x_{1} = 2$ , $y_{1} = 0$ , $x_{2} = 0$ and $y_{2} = 2$ , then:

Slope

$m = \frac{2-0}{0-2}$

$m = -1$

x-Intercept

$b = y_{1} - m\cdot x_{1}$

$b = 0 -(-1)\cdot (2)$

$b = 2$

The equation of the line in the xy-plane is $y = -x+2$ or $x + y = 2$ , which is equivalent to $3\cdot x + 3\cdot y = 6$ .

yz-plane (0, 2, 0) and (0, 0, 3)

$z = m\cdot y + b$

$m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}$

Where:

$m$ - Slope, dimensionless.

$y_{1}$ , $y_{2}$ - Initial and final values for the independent variable, dimensionless.

$z_{1}$ , $z_{2}$ - Initial and final values for the dependent variable, dimensionless.

$b$ - y-Intercept, dimensionless.

If $y_{1} = 2$ , $z_{1} = 0$ , $y_{2} = 0$ and $z_{2} = 3$ , then:

Slope

$m = \frac{3-0}{0-2}$

$m = -\frac{3}{2}$

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$b = z_{1} - m\cdot y_{1}$

$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

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$z = m\cdot x + b$

$m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}$

Where:

$m$ - Slope, dimensionless.

$x_{1}$ , $x_{2}$ - Initial and final values for the independent variable, dimensionless.

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$b = 0 -\left(-\frac{3}{2} \right)\cdot (2)$

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