All categories

4 years ago

Which number is both a perfect square and

Mathematics

1 answer:

lana [24]4 years ago

4 0

C 64
A perfect cube of square is 64

You might be interested in

BRAINLIST and POINTS help

dimaraw [331]

Answer:

Step-by-step explanation:

9*a=63-b=58

because a=7 and b=5

brainliest?

8 0

4 years ago

Read 2 more answers

A wall clock is generally a more precise tool for measuring a length of time than a stopwatch.

Sergio039 [100]

Answer:

False

Step-by-step explanation:

8 0

3 years ago

Jada solved the equation Negative StartFraction 4 over 9 EndFraction = StartFraction x over 108 EndFraction for x using the step

GalinKa [24]

Answer:

Jada should have multiplied both sides of the equation by 108.

Step-by-step explanation:

The question is incomplete. Find the complete question in the comment section.

Given the equation -4/9 = x/108, in order to determine Jada's error, we need to solve in our own way as shown:

Step 1: Multiply both sides of the equation by -9/4 as shown:

-4/9 × -9/4 = x/108 × -9/4

-36/-36 = -9x/432

1 = -9x/432

1 = -x/48

Cross multiplying

48 = -x

x = -48

It can also be solved like this:

Given -4/9 = x/108

Multiply both sides by 108 to have:

-4/9 * 108 = x/108 * 108

-4/9 * 108 = 108x/108

-432/9 = x

x = -48

Jada should have simply follow the second calculation by multiplying both sides of the equation by 108 as shown.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=

BARSIC [14]

<h3>Answer: -2</h3>

======================================================

Work Shown:

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\$

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\$

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

$\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\$

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

$\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\$

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0

3 years ago

Find the radius of a circle having area 616cm square.

OLEGan [10]

Answer:

$\boxed{\sf Radius \ of \ circle = 14 \ cm}$

Given:

Area of circle = 616 cm²

To Find:

Radius of circle

Step-by-step explanation:

$\sf \implies Area \ of \ circle = \pi r^{2} \\ \\ \sf \implies 616 = \frac{22}{7} \times {r}^{2} \\ \\ \sf \implies 616 \times \frac{7}{22} = {r}^{2} \\ \\ \sf \implies 28 \times \cancel{22} \times \frac{7} { \cancel{22}} = {r}^{2} \\ \\ \sf \implies 28 \times 7 = {r}^{2} \\ \\ \sf \implies 196 = {r}^{2} \\ \\ \sf \implies {r}^{2} = 196 \\ \\ \sf \implies r = \sqrt{196} \\ \\ \sf \implies r = \sqrt{ {14}^{2} } \\ \\ \sf \implies r = 14 \: cm$

4 0

3 years ago