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3 years ago

Find the area of the polygon.

Mathematics

2 answers:

monitta3 years ago

7 0

I think the area is 420

Step-by-step explanation:

zhuklara [117]3 years ago

3 0

71yd that’s should be it

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A population, P(t) (in millions) in year t, increases exponentially. Suppose P(9)=16 and P(18)=24. a) Find a formula for the pop

jenyasd209 [6]

Answer: $\frac{32}{3}\left ( 1.0460\right )^t$

Step-by-step explanation:

Given

Population changes exponentially

$P\left ( t\right )=ab^t$

$P\left ( 9\right )=16$ ------1

$P\left ( 9\right )=16=ab^8$

$P\left ( 18\right =24=ab^{18}$ -----2

divide 1 & 2 we get

$\frac{24}{16}=\frac{b^{18}}{b^9}$

$\frac{3}{2}=b^9$

Substitute $b^9$ in 1 we get

$a=\frac{32}{3}$

Thus $P\left ( t\right )=\frac{32}{3}\left ( 1.0460\right )^t$

8 0

3 years ago

Read 2 more answers

Mia and Carlos are hiking on a trail where they start 16 miles apart. If Mia can average 2.3 miles per hour while hiking and Car

kherson [118]

Answer:

I'm not sure but if this is on a calculus paper, this may put you on the right track?

Step-by-step explanation:

definate integral from 2.3 to 2.5 would equal 3.2, i don't know what to do with that information but i hope that helps you.

4 0

2 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Alessandro wrote the quadratic equation –6 = x2 + 4x – 1 in standard form. What is the value of c in his new equation?

Black_prince [1.1K]

Answer:

Step-by-step explanation:

To write it in standard form, we set the equation equal to 0. To do this, we add 6 to each side:

-6+6 = x² + 4x - 1 + 6

0 = x² + 4x + 5

The related function is

y = x² + 4x + 5

The value of c in this function is 5.

7 0

3 years ago

How do you do this question it is kinda difficult to understand

Dafna1 [17]

Speed = slope of the line = 20 / 0.5 = 40 km / hr

Second part will be a line with downward slope from the end of the horizontal line Slop = 20 / 1/3 (=60 km/h) So it will pass through the 65 on horizontal axis.

4 0

3 years ago