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2 years ago

Adiciones racinales 1/7+3/7+2/7

Mathematics

1 answer:

Digiron [165]2 years ago

7 0

6/7 is the awnser for this one

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Tyler’s brother earns a commission. He makes 2.5% of the amount he sells. Last week, he sold $900 worth of shoes. How much was h

lukranit [14]

Answer:

$22.5

Step-by-step explanation:

2.5% of 900

0.025(900)= 22.5

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2 years ago

Find the value of x<br><br> 3(x-4)

antoniya [11.8K]

3x-12 is simplified from parenthesis

7 0

2 years ago

Please help me!!! :(

PSYCHO15rus [73]

Answer:

$15.75x^{10}y^{19}$

Step-by-step explanation:

To find the 5th term in the expansion, we first will need to apply the binomial theorem. I have attached an image of the binomial theorem formula due to not being able to type it.

After applying the binomial theorem and simplifying, you should get:

$512x^{18}y^{27}-576x^{16}y^{25}+288x^{14}y^{23}-84x^{12}y^{21}+\frac{63x^{10}y^{19}}{4}-\frac{63x^8y^{17}}{32}+\frac{21x^6y^{15}}{128}-\frac{9x^4y^{13}}{1024}+\frac{9x^2y^{11}}{32768}-\frac{y^9}{262144}$

Our 5th term here is: $\frac{63x^{10}y^{19}}{4}$ which is equal to $15.75x^{10}y^{19}$

~Hope this helps! Sorry if my answer is confusing at all, it's pretty difficult to explain.~

4 0

2 years ago

there are 75 students on the sapphire team and 1/5 of them are in the band . how many students are in band?

faltersainse [42]

Answer:

Step-by-step explanation:

75/5 = 15

This spilts the 75 students into 5 different groups and each one has 15 which is 1/5 of the students

6 0

1 year ago

Product of two natural numbers p and q is 590 and their hcf is 59. how many set of values of p and q is possible

djverab [1.8K]

$\text{Let the product of two natural numbers p and q is 590, and their HCF is 59}\\
\\
\text{we know that the product of LCM and HCF of any two numbers is equal}\\
\text{to the product of the numbers. that is}\\
\\
\text{HCF}\times \text{ LCM}=p\times q\\
\\
\Rightarrow 59 \times \text{LCM}=590\\
\\
\Rightarrow \text{LCM}=\frac{590}{59}\\
\\
\Rightarrow \text{LCM}=10\\
\\
\text{for any two natural numbers, their Least Common Multiple (LCM) is always}$

$\text{greater than their HCF.}\\
\\
\text{but here we can see that }LCM$

Hence there is no such natural numbers exist.

7 0

2 years ago