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3 years ago

I need help yall. Brainliest for the first person who actually shows their work. 30 pts btw

Mathematics

2 answers:

Oliga [24]3 years ago

7 0

Answer:

1. 24.36

2. Do the same thing (follow the steps in the picture)

Step-by-step explanation:

oee [108]3 years ago

3 0

1. 32 because 15 + 38 = 53 and 90 (the right angle) - 53 is 32

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I have alot of questions and I need help with a practice quiz

Fudgin [204]

ok imma help u just it should not be chem

6 0

2 years ago

Read 2 more answers

7. If U= {2, 4, 6, 8, 10, 12, 14, 16} and A = {6, 10, 14}, what is A'?

Tresset [83]

The compliment of set A that is A' is equal to {2, 4, 8, 12, 16}.

We are given the universal set as:

U = {2, 4, 6, 8, 10, 12, 14, 16}

We are also given that set A is:

A = {6, 10, 14}

We need to find A'

For this, we will subtract set A from the universal set, U.

So, we get that:

A' = U - A

A' = {2, 4, 6, 8, 10, 12, 14, 16} - {6, 10, 14}

Solving the expression, we get that:

A' = {2, 4, 8, 12, 16}

Therefore, we get that, the compliment of set A that is A' is equal to {2, 4, 8, 12, 16}.

Learn more about sets here:

brainly.com/question/2166579

#SPJ9

5 0

8 months ago

Find the number to which the sequence {(3n+1)/(2n-1)} converges and prove that your answer is correct using the epsilon-N defini

Nat2105 [25]

By inspection, it's clear that the sequence must converge to $\dfrac32$ because

$\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32$

when $n$ is arbitrarily large.

Now, for the limit as $n\to\infty$ to be equal to $\dfrac32$ is to say that for any $\varepsilon>0$ , there exists some $N$ such that whenever $n>N$ , it follows that

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|$

From this inequality, we get

$\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}$
$\implies|2n-1|>\dfrac5{2\varepsilon}$
$\implies2n-1\dfrac5{2\varepsilon}$
$\implies n\dfrac12+\dfrac5{4\varepsilon}$

As we're considering $n\to\infty$ , we can omit the first inequality.

We can then see that choosing $N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil$ will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that $N\in\mathbb N$ .

6 0

3 years ago

The value V of a Porsche 718 Cayman that is t years old can be modeled by V(t) = 420,000(0.965)t (a) What would be worth the car

nordsb [41]

Answer:

A. V (2) = $ 391,114.50

B. t = 7.2 years (rounding to the next tenth) or approximately 7 years, 2 months and 12 days

Step-by-step explanation:

Given formula:

V(t) = 420,000 * (0.965)^t

A. What would be worth the car′s worth in 2 years?

V(t) = 420,000 * (0.965)^t

We replace t by 2, as follows:

V(2) = 420,000 * (0.965)²

V(2) = 420,000 * 0.931225

V (2) = $ 391,114.50

B. In how many years will the car be worth $325,000?

V(t) = 420,000 * (0.965)^t

We replace V(t) by 325,000, as follows:

325,000 = 420,000 * (0.965)^t

325,000/420,000 = (0.965)^t

0.77381 = (0.965)^t

t = log 0.965(0.7738)

t = log 0.7738/log 0.965

t = 7.2 years (rounding to the next tenth) or approximately 7 years, 2 months and 12 days

0.2 years = 0.2 * 12 = 2.4 months

0.4 months = 0.4 * 30 = 12 days

7 0

2 years ago

Can someone help me please

Andre45 [30]

Answer:

Is there a choice D in this because A, B and C are all incorrect answers

Step-by-step explanation:

6 0

2 years ago