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3 years ago

Which number rounds to 4.3 when rounded to the nearest tenth?

Mathematics

1 answer:

AVprozaik [17]3 years ago

5 0

Answer:

it is the same 4.3

Step-by-step explanation:

this numbers didn't add one number when you round(0,1,2.3.4)

this numbers(5,6,7,8,9), you add one number when you add

4.3 (the three is the tenth)and no number after it so its stell the same

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Alex Ar [27]

Medical practitioners recommend losing around 365 pounds or 730 pounds per year.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

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2 years ago

PLSSS RNNN<br> -5 + x = 10<br> x = 5<br> x = 10<br> x = -15<br> x = 15

Tema [17]

X = 15

Add 5 to each side
10 + 5 = 15

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2 years ago

Read 2 more answers

Priya ran 1 ½ miles today, and Tyler ran 3 ¾ miles today. How many times the length of Tyler's run was Priya's run?

krek1111 [17]

Answer:
2.5
Step-by-step explanation:

6 0

2 years ago

PLZ HELP SOLVE: -4 < k + 3 < 8

vaieri [72.5K]

Hope this was what you’re looking for ^.^

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3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago