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Dennis_Churaev [7]
2 years ago
11

5 plus 5 equals 10 ......................................

Mathematics
2 answers:
Maru [420]2 years ago
6 0

Answer:

yes 5+5=10 any problem

olga nikolaevna [1]2 years ago
5 0
Yes! 5+5 does equal 10. Imagine you had 5 pencils and you just got another 5 pencils. If you add all those pencils together you get 10 :)
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Need pre-cal help. Will mark best answer brainliest
OlgaM077 [116]

so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.


\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.


\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

5 0
3 years ago
Ralph Chase plans to sell a piece of property for $155000. He
Mrrafil [7]

Ralph chase invested $85000 at an interest of 10% and $70000 at an interest of 8%

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depends on other variable while a dependent variable is a variable that depends on other variable.

Let a represent the amount invested at interest of 10% and b represent the amount invested at 8%, hence:

a + b = 155000    (1)

Also:

0.1a + 0.08b = 14100    (2)

From both equations:

a = 85000, b = 70000

Ralph chase invested $85000 at an interest of 10% and $70000 at an interest of 8%

Find out more on equation at: brainly.com/question/2972832

#SPJ1

7 0
2 years ago
The volume of a cylinder of radius r and height h is given by V = π r 2 h. Then the height, in terms of the radius and volume, i
sdas [7]
Hello There!

It is represented as: h = \frac{v }{ \pi  r^{2} }

Hope This Helps You!
Good Luck :) 

- Hannah ❤
5 0
3 years ago
Read 2 more answers
Write a two-column proof.
-Dominant- [34]

Proof:-

In ∆XYZ and ∆VWZ

\because\sf\begin{cases}\sf XZ\cong VZ(Given)\\ \sf YZ\cong WZ(Given)\\ \sf

Hence

∆XYZ\cong(Side-Angle-Side)

7 0
3 years ago
Find the midpoint of the line segment defined by the points: (5, 4) and (−2, 1) (2.5, 1.5) (3.5, 2.5) (1.5, 2.5) (3.5, 1.5)
Setler79 [48]

Answer:

\boxed {\boxed {\sf (1.5 , 2.5)}}

Step-by-step explanation:

The midpoint is the point that bisects a line segment or divides it into 2 equal halves. The formula is essentially finding the average of the 2 points.

(\frac {x_1+x_2}{2}, \frac {y_1+ y_2}{2})

In this formula, (x₁, y₁) and (x₂, y₂) are the 2 endpoints of the line segment. For this problem, these are (5,4 ) and (-2, 1).

  • x₁= 5
  • y₁= 4
  • x₂= -2
  • y₂= 1

Substitute these values into the formula.

( \frac {5+ -2}{2}, \frac {4+1}{2})

Solve the numerators.

  • 5+ -2 = 5-2 = 3
  • 4+1 = 5

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

Convert the fractions to decimals.

(1.5, 2.5)

The midpoint of the line segment is (1.5 , 2.5)

3 0
3 years ago
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