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sweet-ann [11.9K]
3 years ago
14

I will mark you brainlist!

Mathematics
1 answer:
prisoha [69]3 years ago
5 0

Answer:

the answer is b

Step-by-step explanation:

if it is higher the 3 as you stated it would most likely be b because of it being higher than three so prosecss of elimination you would get b

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I need help. this is the problem x^2/3+10=7x^1/3
Arlecino [84]

\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow  (x^{\frac{1}{3}})^2-7x^{\frac{1}
  {3}} +10=0 
\\\;\\
We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be:
\\\;\\
t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0

\it \Leftrightarrow (t-2)(t-5)=0 
\\\;\\
t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8
\\\;\\
t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5  \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125

S = {8; 125}


7 0
3 years ago
. ∆ABC ~ ∆DEF. Since the triangles are similar, the ratio of AB/DE must be the same as the ratio of ?/DF. What segment complete
givi [52]

Answer:

AC

Step-by-step explanation:

∆ABC ~ ∆DEF

AB/DE

∆ABC ~ ∆DEF

AC/DF

Answer: AC

5 0
3 years ago
Steve opened a bank account.he plans to deposit $35 every month.write an equation that models the total amount of money,t,deposi
zimovet [89]
35m because 35 dollars multiplied by m months
7 0
3 years ago
consider the quadratic functio. f(x)=x2-5x+6 what are the values of the coefficients and constant in the function​
eduard

Answer:

1and5 are coefficients6is a constant

Step-by-step explanation:

6 0
3 years ago
Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10
natka813 [3]

ANSWER

( \frac{2}{3} , \frac{1}{5} )

EXPLANATION

The first equation is

\frac{1}{2} x -  \frac{3}{4} y =  \frac{11}{60} ...(1)

The second equation is

\frac{2}{5} x  +  \frac{1}{6} y =  \frac{3}{10} ...(2)

We want to eliminate y, so we multiply the first equation by

\frac{4}{5}

\frac{4}{5}  \times \frac{1}{2} x - \frac{4}{5}    \times \frac{3}{4} y =  \frac{11}{60}  \times  \frac{4}{5}

\frac{2}{5} x - \frac{3}{5} y =  \frac{11}{75} ...(3)

We now subtract equation (3) from (2)

(\frac{2}{3} x  -  \frac{2}{3} x )+ ( \frac{1}{6} y -  -  \frac{3}{5}y ) =(  \frac{3}{10}  -  \frac{11}{75} )

\frac{1}{6} y  +  \frac{3}{5}y  =\frac{3}{10}  -  \frac{11}{75}

\frac{23}{30}y =  \frac{23}{150}

Multiply both sides by

\frac{30}{23}

\implies \:  \frac{30}{23} \times  \frac{23}{30}y=  \frac{23}{150}  \times  \frac{30}{23}

\implies \: y =  \frac{1}{5}

Substitute into the first equation to solve for x .

\frac{1}{2} x -  \frac{3}{4}  \times \frac{1}{5} =  \frac{11}{60}

Multiply to obtain

\frac{1}{2} x -  \frac{3}{20} =  \frac{11}{60}

\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}

\frac{1}{2} x = \frac{1}{3}

Multiply both sides by 2.

2 \times \frac{1}{2} x =2 \times  \frac{1}{3}

x = \frac{2}{3}

The solution is

( \frac{2}{3} , \frac{1}{5} )

5 0
3 years ago
Read 2 more answers
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