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3 years ago

15

PLEASE HELP with these math questions (Please don't answer if you don't know all of them)

1 answer:

Leona [35]3 years ago

5 0

Sqrt (53) = 10 * sqrt (0.53)

0.53 = 64/121
sqrt (0.53) = sqrt (64)/ sqrt (121) = 8/11 = 0.7273
Therefore sqrt (53) = 10 * 0.7273 = 7.27

sqrt (108) = 10 * sqrt (1.08)
sqrt (1.08) = sqrt (676/625) = 26/25 = 1.04
Therefore sqrt (108) = 10 * 1.04 = 10.4

sqrt (128) = 10 * sqrt (1.28)
sqrt (1.28) = sqrt (289/225) = 17/15 = 1.133
Therefore sqrt (108) = 10 * 1.133 = 11.33

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What is the value of this expression when c = -4 and d = 10?

svlad2 [7]

The answer would be B.9

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3 years ago

Read 2 more answers

The diameter of the circle below is 42 cm. Work out the radius of the circle.

creativ13 [48]

Answer:

r = 21 cm

Step-by-step explanation:

The radius of a circle is 1/2 of the diameter

r = 1/2d

r = 1/2(42)

r = 21 cm

8 0

2 years ago

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

5 0

3 years ago

Alice has $90 to spend on

nlexa [21]

Answer:

5

step by step explanation:

if she gets $10 off of her purchase that means she would have a little extra money to another shirt

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3 years ago

Evaluate 7m+2n-8p/n m=4, n=2, and p=1.5

Firlakuza [10]

Steps to solve:

7m + 2n - 8p/n

~Substitute

7(4) + 2(2) - 8(1.5)/2

~Simplify

28 + 4 - 12/2

~Simplify

28 + 4 - 6

~Add

32

~Subtract

26

Best of Luck!

4 0

3 years ago