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8_murik_8 [283]
3 years ago
6

with vertices D(2, 5), E(6, 4), and F(3, 3), is reflected across the line y=x A student determined one of the vertices on the im

age to be (2, –5). Evaluate the student’s answer
Mathematics
1 answer:
pochemuha3 years ago
8 0
The point (2, -5) would not be one of the correct vertices of the reflected shape.

When we reflected a shape across the line y=x, we just replace the x's with y's and y's with x's.

Our new points are (5, 2) (4, 6) and (3, 3). None of the points contain negative numbers.
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A box is being created out of a 15 inch by 10 inch sheet of metal. Equal-sized squares are cutout of the corners, then the sides
ivolga24 [154]

Answer:

Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box  has largest volume.

Step-by-step explanation:

Given that,

A box is being created out of a 15 inches by 10 inches sheet of metal.

The length of the one side of the squares which are cut out of the each corners of the metal sheet be x.

The length of the metal box be = (15-2x) inches.

The width of the metal box be =(10-2x) inches

The height of the metal box be =x inches

Then, the volume of the metal box= length×width×height

                                                         =(15-2x)(10-2x)x cubic inches

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∴ V= 4x³-50x²+15x

Differentiating with respect to x

V'=12x²-100x+15

Again differentiating with respect to x

V''=24x-100

For maximum or minimum value, V'=0

12x²-100x+15=0

Apply quadratic formula x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, here a=12, b= -100 and c=15

x=\frac{-(-100)\pm\sqrt{(-100)^2-4.12.15}}{2.12}

\Rightarrow x=\frac{100\pm\sqrt{9280}}{2.12}

\Rightarrow x=0.1528,8.18

For x= 8.18, The value of (15-2x) and (10-2x) will negative.

∴x=0.1528 .

Now, V''|_{x=0.1528}=24(0.1528)-100

∴At x=0.1528 inch, the volume of the metal box will be maximum.

Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box  has largest volume.

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Step-by-step explanation:

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