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3 years ago

Help mee…pleasseeessss

Mathematics

1 answer:

jek_recluse [69]3 years ago

6 0

The answer would be: 63 < 2j (im so sorry if I’m wrong)

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A rectangular prism with dimensions 5 inches by 13 inches by 10 inches was cut to leave a piece as shown in the image. What is t

vesna_86 [32]

Answer:

Volume of the piece in the picture = 350 cubic inches

Volume of the remaining piece = 300 cubic inches

Step-by-step explanation:

Given question is incomplete without the picture; here is the picture as an attachment.

Volume of the rectangular prism = Length × width × height

V = 5 × 13 × 10

= 650 cubic inches

Now volume of the piece given in the picture = Volume of prism at the base + Volume of the prism with triangular base

= 10×5×1 + $(\text{Area of the triangular base})\times (\text{Height})$

= 50 + $\frac{1}{2}\times (12\times 5)\times 10$

= 50 + 300

= 350 cubic inches

Volume of the other piece = 650 - 350

= 300 cubic inches

5 0

3 years ago

How to do simplest form

Annette [7]

Step-by-step explanation:

A fraction is in simplest form when the top and bottom cannot be any smaller, while still being whole numbers. To simplify a fraction: divide the top and bottom by the greatest number that will divide both numbers exactly (they must stay whole numbers).

$enjoy yourday$ $hopethishelps$

4 0

3 years ago

Read 2 more answers

5 m<br> 5 m<br> 4 m<br> 8 m<br> AREA:

Damm [24]

Answer:

27.5

Step-by-step explanation:

First you divide up the shape so there is a square and a triangle. Then you would solve for a square: 5 times 4 and you get 20. Then the triangle which you would do 8-5=3 (so you can get the base) and do 3*5/2= 7.5. Then add the two answers

5 0

3 years ago

Read 2 more answers

Given f'(x) = (x-4)(4-2x), find the x-coordinate for relative maximum on the graph of f(x)

antiseptic1488 [7]

Answer:

The coordinate of the relative maximum is x=4.

Step-by-step explanation:

Given that the derivative of the function $f(x)$ is $f'(x) = (x-4)(4-2x)$ , the maxima and minima or the critical points can be found where $f'(x)=0,$ that is:

$f'(x) = (x-4)(4-2x)=0.$

The solutions to this equation are $x=4$ and $x=2.$

Now, if the second derivative $f''(x)$ for a function $f(x)$ is negative at a critical point, then the critical point is the relative maximum.

Therefore we want to see at which of the two critical points is $f''(x)$ negative. The second derivative is:

$\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).$

Now $f''(4)$ is $-4$ , and $f''(2)$ is $+4$ , therefore we deduce that the relative maxium is located at $x=4$ , because there the second derivative $f''(x)$ is negative.

4 0

3 years ago

Free poinnttssssssssssssssss and yourr welcomeee

DerKrebs [107]

Answer:

thank youuuuuu

Step-by-step explanation:

have a good day

5 0

3 years ago