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2 years ago

10(8x - 11) = 690 rlly need help

Mathematics

2 answers:

egoroff_w [7]2 years ago

6 0

Answer:x=10

Step-by-step explanation:

You will first multiply (8x-11) to 10 to distribute.

You end up with 80x - 110.

You will then add 110 from 690.

You end up with 800.

You will then divide by 80, which will leave you with x=10.

Papessa [141]2 years ago

3 0

Answer:

x = 10

Step-by-step explanation:

80x - 110 = 690

80x = 800

x = 10

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Given the function g(x) = 6(4)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Mademuasel [1]

We are given the function:

g(x) = 6 (4)^x

Part A.

To get the average rate of change, we use the formula:

average rate of change = [g(x2) – g(x1)] / (x2 – x1)

Section A:

average rate of change = [6 (4)^1 – 6 (4)^0] / (1 – 0) = 18

Section B:

average rate of change = [6 (4)^3 – 6 (4)^2] / (3 – 2) = 288

Part B.

288 / 18 = 16

Therefore the average rate of change of Section B is 16 times greater than in Section A.

<span>The average rate of change is greater between x = 2 to x = 3 than between x = 1 and x = 0 because an exponential function's rate of change increases with increasing x (not constant).</span>

8 0

3 years ago

How do I find the ratio of this? It makes no sense to me

Volgvan

Answer:

$\frac{169}{289}$

Step-by-step explanation:

Given 2 similar figures with ratio of sides = a : b, then

ratio of areas = a² : b²

Here ratio of sides = 52 : 68 = 13 : 17 ← in simplest form, thus

ratio of areas = 13² : 17² = 169 : 289 = $\frac{169}{289}$

5 0

3 years ago

Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

kondaur [170]

$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .