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4 years ago

What is $10 as a percentage of $40

Mathematics

2 answers:

FinnZ [79.3K]4 years ago

6 0

Hi, I believe the answer is 25%
Hope I Helped :)

Hatshy [7]4 years ago

5 0

0.04 not sure ! but i hope its right

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HELP PLEASE!!! Thanks!

Irina18 [472]

Answer:

2) $\frac{\sqrt{77}}{11}$

3) 5√2

Step-by-step explanation:

Simplest radical form of an expression is the expression in radical so that there are no more square roots, cube roots, 4th roots, etc left to find, after rationalising the denominator ( if needed ).

2. Here the given expressions,

$\frac{\sqrt{7}}{\sqrt{11}}$

For rationalising the denominator multiply both numerator and denominator by √11,

$\frac{\sqrt{77}}{11}$

3. given expression,

$\sqrt{50}$

$=\sqrt{25\times 2}$

$=\sqrt{25}\times \sqrt{2}$ $(\because \sqrt{ab}=\sqrt{a}\sqrt{b})$

$=5\sqrt{2}$

8 0

3 years ago

What is 2 2/5×2 7/9 in simplest mixed number form?

Scorpion4ik [409]

Answer:

6 2/3

Step-by-step explanation:

2 2/5 * 2 7/9

Change each number to an improper fraction

2 2/5 = (5*2+2)/5 = 12/5

2 7/9 = (9*2+7) = 25/9

12/5 * 25/9

Rewriting

25/5 * 12/9

5/1 * 4/3

20/3

Now changing to a mixed number

3 goes into 20 6 times with 2 left over

6 2/3

7 0

4 years ago

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$