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4 years ago

8

What is a perfect square

2 answers:

Gelneren [198K]4 years ago

7 0

The square of an integer is called a perfect square. For example, 49 is a perfect square because 7² = 49. When a radicand is not a perfect square, you can estimate the square root of the radicand.

Troyanec [42]4 years ago

6 0

4 equal sides, 4 right angles, 2 pairs of parallel and opposit sides I think

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How many ounces of meat are used to make a 1/4-In hamburger?

Liono4ka [1.6K]

The answer is to 1/4 hamburger is 4 oz

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3 years ago

CAN SOMEONE HELP ME PLEASEEE

Tom [10]

Answer:

a. 45 degrees

b. 315 degrees

c. 8π

d. π

e. 7π

f. 16π

g. 2π

6 0

3 years ago

Limit question: lim x-->pi ((e^sinx)-1)/(x-pi)

aleksandr82 [10.1K]

$\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}$

Notice that if $f(x)=e^{\sin x}$ , then $f(\pi)=e^{\sin\pi}=e^0=1$ . Recall the definition of the derivative of a function $f(x)$ at a point $x=c$ :

$f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So the value of this limit is exactly the value of the derivative of $f(x)=e^{\sin x}$ at $x=\pi$ .

You have

$f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1$

4 0

4 years ago

please help :{ brainliest if you can get it right and I need you to prove why your answer is correct, thanksss

beks73 [17]

Answer:

The Value of $a=\frac{15}{4}$ .

Step-by-step explanation:

We have Named the figure please find the attachment for your reference.

Given:

PR = y

QR = a

RS = b

PS = z

PQ = x

QS = 15

∠P = 90°

∠R = 90°

∠Q = 60°

∠S = 30°

We need to find the Value of 'a'.

Solution:

Now we know that:

In Δ PQS

∠P = 90°

∠S = 30°

Now we know that;

$sin\ \theta = \frac{opposite\ side}{Hypotenuse}$

$sin \ S= \frac{PQ}{QS}$

Substituting the given values we get;

$sin\ 30\°=\frac{x}{15}$

Now we know that;

$sin\ 30\° = \frac12$

So we can say that;

$\frac{1}{2}=\frac{x}{15}\\\\x=\frac{15}{2}$

Now In Triangle PQR.

∠R = 90°

∠Q = 60°

So we can say that;

$Cos \theta = \frac{adjacent \ Side}{Hypotenuse}\\$

$Cos\ Q = \frac{QR}{PQ}$

Substituting the given values we get;

$cos 60\°= \frac{a}{x}$

Now we know that;

$cos 60\°= \frac12$

$x=\frac{15}{2}$

So substituting the values we get;

$\frac{1}{2}=\frac{a}{\frac{15}{2}}$

By Using Cross Multiplication we get;

$a= \frac{1}{2}\times\frac{15}{2}\\\\a=\frac{15}{4}$

Hence The Value of $a=\frac{15}{4}$ .

6 0

3 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

3 years ago