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2 years ago

Can 5/12 be simlified

Mathematics

2 answers:

8_murik_8 [283]2 years ago

6 0

No 5/12 is already in its simplest form, and cannot be broken down further.

nikklg [1K]2 years ago

4 0

It's already in simplest form.
So you wouldn't be able to simplify it further.

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Can someone help me? I’ll reward points + brainalist

Setler79 [48]

Answer:

d) A = π( $9^{2}$ )

Step-by-step explanation:

area = π $r^{2}$

diameter = 18

radius (half of diameter) = 9

area = π $9^{2}$

6 0

2 years ago

Sam is a waiter at a local restaurant where he earns wages of $7per hour. Sam figures that he also earns about $5 in tips for ea

Mnenie [13.5K]

There aren't any functions listed to choose from, but it should look something like the following:

E= total earnings that day
h= # hours worked that day
n= # people served that day

E= 7h + 5n

substitute 6 hours in for h

E= 7(6) + 5n

E= 42 + 5n OR E= 5n + 42

Hope this helps! :)

8 0

3 years ago

Solve the equation using inverse operations. Check your solutions. In your final answer, include all of your work.

REY [17]

Answer:

$x=\sqrt{10}$

Step-by-step explanation:

The given equation is :

$5-2x^2=-15$

We need to solve it using inverse operations.

Subtract 5 from both sides of the equation.

$5-2x^2-5=-15-5\\\\-2x^2=-20\\\\2x^2=20\\\\x^2=10\\\\x=\sqrt{10}$

So, the solution of the given equation is $x=\sqrt{10}$ .

4 0

2 years ago

If the equation; (3x)² + {27 × (3)^1/k - 15}x + 4 = 0 has equal roots find k

Dima020 [189]

Step-by-step explanation:

$\green{\large\underline{\sf{Solution-}}}$

Given quadratic equation is

$\rm :\longmapsto\:\rm \: {(3x)}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0$

can be rewritten as

$\rm :\longmapsto\:\rm \: {9x}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0$

Concept Used :- 

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Let's Solve this problem now!!!

On comparing with quadratic equation ax² + bx + c = 0, we get

$\red{\rm :\longmapsto\:a = 9}$

$\red{\rm :\longmapsto\:b = 27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15}$

$\red{\rm :\longmapsto\:c = 4}$

Since, Discriminant, D = 0

$\rm \implies\: {b}^{2} - 4ac = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 4 \times 4 \times 9 = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 144 = 0$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = 144$

$\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = {12}^{2}$

$\rm \implies\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = \: \pm \: 12$

Case - 1

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = - 12$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = - 12 + 15$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 3$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{9}$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{ {3}^{2} }$

$\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{ - 2}$

$\rm \implies\:\dfrac{1}{k} = - 2$

$\bf\implies \:k \: = \: - \: \dfrac{1}{2}$

So, option (b) is Correct.

Case - 2

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = 12$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 12 + 15$

$\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 27$

$\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 1$

$\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{0}$

$\rm \implies\:\dfrac{1}{k} =0$

which is not possible.

7 0

1 year ago

What is the simplified form of this expression? (8x − 7) + (-2x − 9) − (4x − 3) A. 2x – 13 B. 2x − 19 C. 2x − 5 D. 2x + 1

Yanka [14]

Answer:

A. 2x-13

Step-by-step explanation:

5 0

2 years ago