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3 years ago

Multiply (Please help)(x+4)²

Mathematics

1 answer:

frosja888 [35]3 years ago

7 0

Answer:

${(x + 4)}^{2} \\ (x + 4)(x + 4) \\ {x}^{2} + 4x + 4x + 16 \\ {x}^{2} + 8x + 16$

I hope I helped you^_^

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The surface area of the prism is ______ square units. All measurements in the image below are in units. (Input whole number only

White raven [17]

To find the area of a triangle, use 1/2 x b x h. 3 x 4 = 12 1/2 of 12 = 6

There are two triangles, each of these are 6 6 x 2 = 12

The bottom shape is 3.5 x 3 3.5 x 3 = 10.5

The front shape is 3.5 x 5 3.5 x 5 = 17.5

The back shape is 3.5 x 4 3.5 x 4 = 14

Add them all together: 14 + 17.5 + 10.5 + 12 = 54

Your answer is 54!

7 0

3 years ago

Giving brainliest for correct answer!

Mamont248 [21]

Answer:

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30 21

Step-by-step explanation:

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3 years ago

Rectangle A has a length of 12 and width of 8. Rectangle B has a length of 15 and a width of 10.

navik [9.2K]

Answer:

rectangles are similar figures, thus if scaled copies of each other then the ratios of corresponding sides must be equal

compare ratios of lengths and widths

rectangles A and B

k = = ← ratio of lengths

k = = ← ratio of widths

scale factors are equivalent, hence rectangle A is a scaled copy of B

rectangles C and B

k = = ← ratio of lengths

k = = ← ratio of width

scale factors (k ) are not equal, hence C is not a scaled copy of B

rectangles A and C

k = = ← ratio of lengths

k = ← ratio of widths

the scale factors are not equal hence A is not a scaled copy of C

Step-by-step explanation:

6 0

3 years ago

Find the length of the curve y = integral from 1 to x of sqrt(t^3-1)

Arlecino [84]

$y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}$

Now the arc length over an arbitrary interval $(a,b)$ is

$\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx$

But before we compute the integral, first we need to make sure the integrand exists over it. $x^{3/2}$ is undefined if $x$ , so we assume $a\ge0$ and for convenience that $a$ . Then

$\displaystyle\int_a^bx^{3/2}\,\mathrm dx=\frac25x^{5/2}\bigg|_{x=a}^{x=b}=\frac25\left(b^{5/2}-a^{5/2}\right)$