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DedPeter [7]
3 years ago
15

What is -2/3x+1/3+x=2/5? thanks.

Mathematics
1 answer:
OverLord2011 [107]3 years ago
3 0
-2/3x +1/3 +x = 2/5
⇒ 1/3x + 1/3 = 2/5
⇒ 1/3x = 2/5 -1/3
⇒ x/3 = 1/15
⇒ 15x = 3
⇒ x = 3/15
⇒ x = 1/5
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How many ounces of a 15​% alcohol solution must be mixed with 4 ounces of a 20​% alcohol solution to make a 17​% alcohol​ soluti
sukhopar [10]

Answer:

6 ounces

Step-by-step explanation:

Let x = the number of ounces of the 15% alcohol solution needed to be mixed with the 4 ounces of 20% alcohol solution to obtain the 17% alcohol solution.

After converting the percentages to their decimal equivalents, set up the necessary equation as follows:

0.15x+0.20(4) = (x+4)(0.17) Simplify and solve for x.

0.15x+0.8 = 0.17x+0.68 Subtract 0.15x from both sides.

0.8 = 0.02x+0.68 Now subtract 0.68 from both sides.

0.12 = 0.02x Finally, divide both sides by 0.02

x = 6

You need to mix 6 ounces of 15% alcohol solution with 4 ounces of 20% alcohol solution to obtain 10 (x+4)ounces of 17% alcohol solution.

7 0
2 years ago
Read 2 more answers
Solve the equation.<br><br> 3(x−9)=30
a_sh-v [17]

Answer:

x = 19

Step-by-step explanation:

3(x - 9) = 30

Divide both sides by 3 to isolate the binomial.

x - 9 = 10

Add 9 to both sides to isolate x.

x = 19

Check your answer by plugging x = 19 back into the equation.

3(x - 9) = 30

3(19 - 9) = 30

Subtract.

3(10) = 30

Multiply.

30 = 30

Your answer is correct.

Hope this helps!

3 0
3 years ago
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What's the simplified form of this expression? 3-[-2/3 x (7 +2) =11]
svetoff [14.1K]

Answer:

x = 4/3

Step-by-step explanation:

7 0
3 years ago
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?
sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have V=IR^{2} and let H be the subset of V of all points on the line

4x+3y=12

We need to find if H is a subspace of the vector space V.

In IR^{2} all the possibilities for own subspace of the vector space IR^{2} are :

  • IR^{2} itself.
  • The vector 0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]
  • All lines in IR^{2} that passes through the origin  (  0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]  )

We know that H is the subset of IR^{2} of all points on the line 4x+3y=12

If we look at the equation, the point \left[\begin{array}{c}0&0\end{array}\right] doesn't verify it because :

4x+3y=12\\4(0)+3(0)=12\\0=12

Which is an absurd. Therefore, H doesn't contain the origin (and H is a line in IR^{2}). Finally, it can't be a vector space of V=IR^{2}

8 0
3 years ago
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