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3 years ago

47 over 5 as percentage

Mathematics

1 answer:

aleksklad [387]3 years ago

6 0

940% because 47 divided by 5 is 9.4.

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The approximate weights of two animals are 4.23 x 103 lbs. and 8.7 x 103 lbs. find the total weight of the two animals. write th

vagabundo [1.1K]

The total weight of two animals in scientific notation is 1.3 x 10⁴ lbs.

Scientific Notation:

A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10.

For example,

650,000,000 can be written in scientific notation as 6.5 ✕ 10⁸.

Given,

The approximate weights of two animals are 4.23 x 10³ lbs. and 8.7 x 10³ lbs.

Here we need to find the total weight of the two animals and we need to write the final answer in scientific notation with the correct number of significant digits.

The total weight of the two animals is calculated as

Total weight = Animal 1 + Animal 2

So,

Total weight = 4.23 x 10³ lbs + 8.7 x 10³ lbs.

Take 10³ lbs as the common term,

Then we get,

Total weight = (4.23 + 8.7) x 10³ lbs.

Total weight = 12.93 x 10³ lbs.

It can we rewritten as,

Total weight = 1.293 x 10⁴ lbs.

When we round off it, then we get,

Total weight = 1.3 x 10⁴ lbs.

Hence, the total weight of the two animals is 1.3 x 10⁴ lbs.

To know more about Scientific Notation here.

brainly.com/question/18073768

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8 0

1 year ago

If -2m+3=m-12 what is the value of 10m

ra1l [238]

The answer is 50 because m=5

6 0

3 years ago

WILL GIVE BRAINLIEST <br> AND 20 POINTS

fiasKO [112]

10 is the answer so yea add me on IG
xbl.bodey for free mods

4 0

4 years ago

I need help fast! Answer and explanation please! (see attachment) Solve for x, show your work.

Inessa05 [86]

Answer:

x = -5/2

Step-by-step explanation:

3 ^ ( 4x-5) = (1/27) ^( 2x+10)

Rewriting 1/27 as 3^-3

3 ^ ( 4x-5) = (3^-3) ^( 2x+10)

We know that a^b^x = a^(b*c)

3 ^ ( 4x-5) = (3) ^(-3*( 2x+10))

3 ^ ( 4x-5) = (3) ^( -6x-30)

The bases are the same so the exponents are the same

4x-5 = -6x-30

Add 6x to each side

10x -5 = -30

Add 5 to each side

10x -5+5 = -30+5

10x = -25

Divide each side by 10

10x/10 = -25/10

x = -5/2

4 0

4 years ago

Read 2 more answers

GIVING OUT BRAINLIEST TO WHOEVER GETS ALL OF THEM RIGHT

Thepotemich [5.8K]

Answer:

4) $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$ is equivalent to $\frac{1}{x-3}$ for all $x \ne 3$ . (Answer: None)

7) $\frac{x^{2}+3\cdot x -4}{x+4}$ is equivalent to $x - 1$ . (Answer: None)

8) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$ is equivalent to $\frac{4}{3\cdot a^{3}}$ for all $a\ne 0$ . (Answer: A)

Step-by-step explanation:

We proceed to simplify each expression below:

4) $\frac{x}{7\cdot x +x^{2}}$

(i) $\frac{x}{7\cdot x +x^{2}}$ Given

(ii) $\frac{x}{x\cdot (7+x)}$ Distributive property

(iii) $\frac{1}{7+x} \cdot \frac{x}{x}$ Distributive property

(iv) $\frac{1}{7+x}$ Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$7+x = 0$

$x = -7$

Hence, we conclude that $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$

(i) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ Given

(ii) $\frac{x^{3}\cdot (-14)}{x^{3}\cdot (1-5\cdot x)}$ Distributive property

(iii) $\frac{x^{3}}{x^{3}} \cdot \left(-\frac{14}{1-5\cdot x} \right)$ Distributive property

(iv) $-\frac{14}{1-5\cdot x}$ Commutative property/Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$1-5\cdot x = 0$

$5\cdot x = 1$

$x = \frac{1}{5}$

Hence, we conclude that $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$

(i) $\frac{x+7}{x^{2}+4\cdot x - 21}$ Given

(ii) $\frac{x+7}{(x+7)\cdot (x-3)}$ $x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})$

(iii) $\frac{1}{x-3}\cdot \frac{x+7}{x+7}$ Commutative and distributive properties.

(iv) $\frac{1}{x-3}$ Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$x-3 = 0$

$x = 3$

Hence, we conclude that $\frac{x+7}{x^{2}+4\cdot x - 21}$ is equivalent to $\frac{1}{x-3}$ for all $x \ne 3$ . (Answer: None)

7) $\frac{x^{2}+3\cdot x -4}{x+4}$

(i) $\frac{x^{2}+3\cdot x -4}{x+4}$ Given

(ii) $\frac{(x+4)\cdot (x-1)}{x+4}$ $x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})$

(iii) $(x-1)\cdot \left(\frac{x+4}{x+4} \right)$ Commutative and distributive properties.

(iv) $x - 1$ Existence of additive inverse/Modulative property/Result

Polynomic function are defined for all value of $x$ .

$\frac{x^{2}+3\cdot x -4}{x+4}$ is equivalent to $x - 1$ . (Answer: None)

8) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$

(i) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$

(ii) $\frac{4}{3\cdot a^{3}}$ $\frac{a}{b}\cdot \frac{c}{d} = \frac{a\cdot b}{c\cdot d}$ /Result

Rational functions are undefined when denominator equals 0. That is:

$3\cdot a^{3} = 0$

$a = 0$

Hence, $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$ is equivalent to $\frac{4}{3\cdot a^{3}}$ for all $a\ne 0$ . (Answer: A)

6 0

3 years ago