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2 years ago

What is the area of rectangle ABCD in square units? , I’ll mark brainliest

Mathematics

1 answer:

tiny-mole [99]2 years ago

3 0

Answer:

What is the area of rectangle ABCD in square units?

, I’ll mark brainliest

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Find AB Round to the nearest tenth.

lianna [129]

Answer:4.1

Step-by-step explanation:

c=AB

C=26°

B=122°

b=8

Using sine rule

c/sinC = b/sinB

c/sin26=8/sin122

Sin26=0.4384

Sin122=0.8480

c/0.4384=8/0.8480

Cross multiply

c x 0.8480=8 x 0.4384

c x 0.8480=3.5072

Divide both sides by 0.8480

(c x 0.8480)/0.8480=3.5072/0.8480

c=4.1

7 0

2 years ago

Find the standard equation of the parabola that satisfies the given conditions. Also, find the length of the latus rectum of eac

lakkis [162]

Answer:

The standard parabola

y² = -18 x +27

Length of Latus rectum = 4 a = 18

Step-by-step explanation:

Explanation:-

Given focus : (-3 ,0) ,directrix : x=6

Let P(x₁ , y₁) be the point on parabola

PM perpendicular to the the directrix L

SP² = PM²

(x₁ +3)²+(y₁-0)² = $(\frac{x_{1}-6 }{\sqrt{1} } )^{2}$

x₁²+6 x₁ +9 + y₁² = x₁²-12 x₁ +36

y₁² = -18 x₁ +36 -9

y₁² = -18 x₁ +27

The standard parabola

y² = -18 x +27

Length of Latus rectum = 4 a = 4 (18/4) = 18

5 0

2 years ago

Tyler’s brother earns $14 per hour. The store offers him a raise... an 8% increase per hour. After the raise, how much will Tyle

GaryK [48]

He will make $15.12 a hour from the raise.

3 0

2 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

motel clerk counts his $1 and $10 bills at the end of a day. He finds that he has a total of 57 bills having a combined monetary

madreJ [45]

You can use a matrix equation to discover that the motel clerk has 45 $1 bills and 12 $10 bills.

4 0

3 years ago