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3 years ago

Please help me I am having a hard time understanding this !

Mathematics

1 answer:

FromTheMoon [43]3 years ago

6 0

Ududd8rhr8he8 uh egigeigs8heiheoyoueohrdohpduepup0eu0ie0eie

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Plz help ill give u brainlist

Anna35 [415]

Sense this would be known to be a 'prism', we would them do the following key point:

$l*w*h$

Therefore, we would then multiply to following numbers which would be above.

$(5*4*3=60)$

Question #2.

The reason to why we would us 'square units' to answer in a area format would be because by doing so, it would then describe the surface of the area. And by knowing what you are measuring something by would be a whole lot easier.

8 0

3 years ago

Read 2 more answers

kim practices soccer each day. she made a chart of the number of shots she made and the goals she scored. find the equation that

Sedaia [141]

Answer:4x

Step-by-step explanation:

8 0

2 years ago

Simplify 8(7x – 1) + 4(x + 5).

Elina [12.6K]

8(7x-1)+4(x+5)

multiply; 8 x 7=

56x + -1 + 4x +5

reorder the terms;

-1 + 5 + 56x + 4x

Combine like terms; -1 + 5 = 4

4 + 56x + 4x

Combine like terms again; 56x + 4x = 60x

4 + 60x

4 0

3 years ago

Read 2 more answers

What are the zeros of the function f(x) = (x + 5)(2x-3)?

ohaa [14]

The answer would be C done this yesterday

6 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago