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3 years ago

I need help as soon as possible

Mathematics

2 answers:

mote1985 [20]3 years ago

5 0

The sin ratio is the side opposite the reference angle over the hypotenuse. If Z is our reference angle, then the side opposite it has a length of 32. The hypotenuse has a length of 40. Therefore, the sin of Z = 32/40. We will reduce that to sin Z = 4/5. First choice above.

RoseWind [281]3 years ago

3 0

For Sinz you have to do
$\frac{32}{40}$
so answer is
$\frac{4}{5}$
i think and what is your school level?

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Y_Kistochka [10]

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3 years ago

Arah went to the grocery store and bought 4 loaves of bread and 1 gallon of milk for $12. The next week, Sarah bought 2 loaves o

sergejj [24]

Answer:

Here, b represents one loaf of bread and m represents the one gallon of milk.

As per the statement:

Arah went to the grocery store and bought 4 loaves of bread and 1 gallon of milk for $12.

⇒ $4b+m = 12$

It is also given that the next week, Sarah bought 2 loaves of bread and 3 gallons of milk for $13.50.

⇒ $2b+3m = 13.50$

Then; system of equation :

$4b+m = 12$ .....[1]

$2b+3m = 13.50$ .....[2]

Solve for b and m using above system of equations.

Multiply equation [2] by 2 we get;

$4b+6m = 27$ .....[3]

Subtract equation [1] from [3] we get;

$4b+6m - 4b -m= 27-12$

Combine like terms;

$5m= 15$

Divide both sides by 5 we get;

m = $3

Substitute the value of m in equation [1] we get;

$4b+3 = 12$

Subtract 3 from both sides we get;

$4b=9$

Divide both sides by 4 we get;

b = $2.25

Therefore, cost of one loaf of bread (b) and one gallon of milk (m) are:

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3 years ago

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Nat2105 [25]

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

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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

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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 $N\in\mathbb N$ .