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3 years ago

Need help again pls. Kind of urgent

Mathematics

1 answer:

Ber [7]3 years ago

8 0

You know we can't draw it, but I will give you all information so that to enable you draw this parabola

y = (x-4)² +3.

1-The coefficient of x =1 so it's positive, the parabola passe by a minimum or opens upward
2- It doesn't intercepts the x-axis since it has no zero value
3- its shifted 3 units upward & shifted 4 units to the right
4- its axis of symmetry is x=4 & its minimum is at (-3 , 3)
Hope this will help

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Answer:

Step-by-step explanation

Hello!

3x/4 + 6=42

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Find the zeros or roots for the quadratic expression 2^2−8−10. Must show all work to receive credit

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x = -1,5

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3 years ago

The new cylinder can has radius of 5cm and a height of 9cm how many square centimeters of aluminum will it take to make this new

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Answer: $282.85\ cm^2$

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The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

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Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

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$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

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