What did the ancestral puebloans do to make the growing need for housing

1 answer:

8 0

Answer: How did the Ancestral Puebloans adapt their housing to the building materials that were available in the arid Southwest? They built some houses of stone blocks against canyon walls and built freestanding villages of stone and adobe brick.

Step-by-step explanation:

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What is the volume of a prism if the base equals 176 cm square and the width is 4 cm

Nimfa-mama [501]

704 ( I'm assuming this is rectangular prism )

Follow this formula:

width x length x height = volume

Base is height x length so half of the problem is solved. Next just multiply that by the width and you found your answer.

5 0

2 years ago

Solve −5x = 15. (1 point) −3 3 10 20

Luba_88 [7]

Remember you can do anything to an eqautoin as long as you do oit to both sides

-5x=15
try to get 1x by itself

remember
(ax)/a=x when a=a

so
-5x=15
get x
divide both sides by -5
remember to flip sign
(-5x)/(-5)=15/(-5)
x=-3

answer is first one
x=-3

8 0

2 years ago

Read 2 more answers

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$