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3 years ago

Roman food was completely different from the food we eat today. True False

2 answers:

8 0

<span>Roman food was completely different from the food we eat today.
Answer : true, in many ways romans ate different from us, you would find tht many delacacies today are considered discusting today</span>

Julli [10]3 years ago

3 0

The answer is True..

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Which statement describes an example of gerrymandering?

enyata [817]

Answer:

Explanation:

C is almost an exact definition of what Gerrymandering is. It is the redistribution of populations within an area that will favor one political party or one political candidate.

7 0

3 years ago

Read 2 more answers

The table above gives values of f, f’, g, and g’ at selected values of x. If h(x) = f(g(x)), then h’(1) =

Anit [1.1K]

Answer:

h'(-1) = 5

Explanation:

Given - The table above gives values of f, f’, g, and g’ at selected values

of x.

x f(x) f'(x) g(x) g'(x)

-1 6 5 3 -2

1 3 -3 -1 2

3 1 -2 2 3

To find - If h(x) = f(g(x)), then h’(1) = ......?

Proof -

Given that,

h(x) = f(g(x))

⇒h'(x) = f'(g(x))

⇒h'(1) = f'(g(1))

Now,

g(1) = -1

⇒f'(-1) = 5

⇒h'(-1) = 5

3 0

3 years ago

Read 2 more answers

The image shows a map of the number of operating spindles in 1840. A map of the number of operating spindles in the U S in 1840.

VashaNatasha [74]

Answer:

it may be c bc transboration did improve

Explanation:?

6 0

3 years ago

If y = 3x3 – 2x and dx/dt=3, find dy/dt when x = –2. Give only the numerical answer. For example, if dy/dt = 4, type only 4.

ludmilkaskok [199]

-6 dy\dx multiplied by dx\dt will give you dy\dt and dy\dx=-2 and dx\dt=3

5 0

4 years ago

I can’t figure this problem out

Natalka [10]

Answer:

Explanation:

Alright so the way to do this is to use properties of integrals to make our life easier.

So we have:

$\int\limits^4_1 {(3f(x)+2)} \, dx$

So lets break this up into two different integrals that represent the same area.

$\int\limits^4_0 {f(x)} \, dx - \int\limits^1_0 {f(x)} \, dx = \int\limits^4_1 {f(x)} \, dx$

Lets think about what is going on up there. The integral from four to zero gives us the area under the curve of f(x) from four to zero. If we subtract this from the integral from one to zero (the area under f from one to zero) we are left with the area under f from four to one! Hence:

$\int\limits^4_1 {f(x)} \, dx$

But since we have these values we can say that:

-3 - 2 = -5

Which means that $\int\limits^4_1 {f(x)} \, dx$ = -5

So now we can evaluate $\int\limits^4_1 {(3f(x)+2)} \, dx$

Lets first break up our integrand into two integrals

$\int\limits^4_1 {(3f(x)+2)} \, dx$ = $3\int\limits^4_1{f(x)} \, dx + 2\int\limits^4_1 {} \, dx$

Now we can evaluate this:

We know that $\int\limits^4_1 {f(x)} \, dx$ = -5

So:

$3(-5)+2[x]$ where x is evaluated at 4 to 1 so

-15 + 2(3)

So we are left with -15 + 6 = -9

5 0

3 years ago