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Sophie [7]
2 years ago
14

One person can do a certain job in fifteen minutes, and another person can do the same job in ten minutes. How many minutes will

they take to do the job together?
Mathematics
2 answers:
8_murik_8 [283]2 years ago
8 0
It would take 5 because they could do it twice as fast
MissTica2 years ago
4 0

Answer:

They will take 6 minutes to do the job together.

Step-by-step explanation:

Consider the provided information.

It is given that the time taken by one person to do a job is 15 minutes.

The time taken by another person to do the same job is 10 minutes.

Let T represent the time taken when they both work together.

Therefore the required equation is:

\frac{1}{T}=\frac{1}{15}+\frac{1}{10}

\Rightarrow\frac{1}{T}=\frac{2+3}{30}

\Rightarrow\frac{1}{T}=\frac{5}{30}

\Rightarrow\frac{1}{T}=\frac{1}{6}

\Rightarrow T=6

Hence, they will take 6 minutes to do the job together.

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Simplify the expression. 12x^-6 y^10 times 3x^7 y
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We need to simplify the expression:

12x^{-6}y^{10} \times (3x^{7}y)

Now, we know that we can resolve the exponents of the variables with the like terms only and we can multiply the coefficients independently:

Now,  

12x^{-6}y^{10} \times 3x^{7}y=(12 \times 3)\times (x^{-6}\times x^{7})\times (y^{10}\times y)

On simplifying the above expression we get:

36\times x^{(-6+7)}\times y^{(10+1)}

=36 \times x^{1}\times y^{11}

=36 \times x \times y^{11}

=36xy^{11}

So the simplified form of the expression 12x^{-6}y^{10} \times 3x^{7}y=36xy^{11}.

7 0
2 years ago
Mr. Altamirano opens an account with a simple interest on an account with $1000 at 3% annually for 9 months. How much interest d
gladu [14]

Answer:

1022.50

Step-by-step explanation:

1000*.03= 30. (12mos interest)

30/4=7.50 quarterly interest

7.50x3 (9mos)= 22.50 interest

7 0
3 years ago
Find which term in the geometric sequence 1,3,9,27,... is the first to exceed 7,000.
Zielflug [23.3K]

The common ratio between terms is 3, so the sequence has general n-th term

a_n=3^{n-1}

for n\ge1. The term exceeds 7000 when

3^{n-1}>7000\implies n-1>\log_37000\implies n>1+\log_37000\approx9.06

which means the first time a_n exceeds 7000 occurs when n=10. Indeed,

a_{10}=3^{10-1}=19,683

while the previous term would have been

a_9=3^{9-1}=6561

8 0
3 years ago
Solve step by step solution then only i can do it plxx ​
Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So \triangle ABC and \triangle ABD Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
  • First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

  • We need to find base=b

According to Pythagoras thereon

{\boxed{\sf b^2=h^2-p^2}}

  • Substitutethe values

\longrightarrow\sf b^2=10^2-p^2

\longrightarrow\sf b={\sqrt {10^2-8^2}}

\longrightarrow\sf b={\sqrt{100-64}}

\longrightarrow\bf b={\sqrt {36}}

\longrightarrow\sf b=6

\therefore\overline{BC}=6cm

  • BD=BC+CD

\longrightarrowBD=9+6

\longrightarrowBD=15cm

  • Now in \triangle ABD

Perpendicular=p=8cm

Base =b=15cm

  • We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

{\boxed {\sf h^2=p^2+b^2}}

  • Substitute the values

\longrightarrow\sf h^2=8^2+15^2

\longrightarrow\sf h={\sqrt {8^2+15^2}}

\longrightarrow\sf h={\sqrt {64+225}}

\longrightarrow\sf h={\sqrt {289}}

\longrightarrow\sf h=17cm

\therefore{\underline{\boxed{\bf x=17cm}}}

3 0
2 years ago
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