All categories

3 years ago

Can anyone help me? Brainlist if right! Thx :)

Mathematics

1 answer:

tamaranim1 [39]3 years ago

4 0

Answer:

odd: 4/7

not 3: 6/7

4 or 5: 2/7

spinner spun 2: there both 1/7 i think im

only in 7th grade ;-;

Step-by-step explanation: well there are 7 spaces so the probabilty

of it landing on an odd space on the wheel is how many odd numbers there are over 7 . and there is only 1 three so the chance that it will not land on 3 is 6/7. there is a 2/7 chance it will land on 4 or 5. and im not to sure abt the last one but like i said im only in the 7th grade i am taking p classes tho so the rest should be right. :D

You might be interested in

Find the value of x.<br> 75°<br> 3x°<br> X =<br> degrees

kipiarov [429]

Explanation:
Sum of all angles= 180 degrees

So,
3x + 75= 180
3x= 180 - 75
3x= 105
x= 105/3
x= 35

Answer= 35 degrees

5 0

2 years ago

Dedra used coupons 6 times at the store last month.

AVprozaik [17]

Answer:

The mean is 42, rounded to the nearest cant would be 42.00. IF this helped subscribe to Amiredagoat Yt

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Whats 2+2+9+7+10+67+89+572+53+4=?

Sav [38]

Step-by-step explanation:

815..............…......…

7 0

3 years ago

Read 2 more answers

I gotta go stinky in the toilet. Do i. A. Go make stinky in the toilet now and do my assignment later, or B. go stinky in my pan

lbvjy [14]

Answer:

go stinky in the toilet now and homework later

9 0

3 years ago

Read 3 more answers