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Dafna11 [192]
3 years ago
12

Please I need help badly and quickly, pls do not answer if you just want the points. I need an accurate answer for this. Cheers

Mathematics
1 answer:
natita [175]3 years ago
7 0

Answer

Q11 is 50,3

Q12, 23,19

Step-by-step explanation:

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Need help with us question
lina2011 [118]

Answer:

17. 20  

18. 50

19. 50

20. 70

21. 110

22. 160

Step-by-step explanation:

I did this assignment already :)

7 0
3 years ago
What is the solution to the expression 12÷4+(1/3)²⋅9⋅3 ?
krok68 [10]

Answer:

6

Step-by-step explanation:

STEP

1

:

           1

Simplify   —

           3

The equation at the end of step

1

:

 12       1

 —— +  (((—)2) • 9) • 3)

 4        3

STEP

2

:

Equation at the end of step

2

:

 12       1

 —— +  ((—— • 9) • 3)

 4       32

STEP

3

:

Canceling Out:

3.1      Canceling out  32 as it appears on both sides of the fraction line

Equation at the end of step

3

:

 12    

 —— +  (1 • 3)

 4    

STEP

4

:

           3

Simplify   —

           1

Equation at the end of step

4

:

 3 +  3

STEP

5

:

Pulling out like terms

5.1      Pull out     3

After pulling out, we are left with :

     3 • ( 1 - (-1) ))

Final result :

 6

5 0
2 years ago
Read 2 more answers
Show that the line 4y = 5x-10 is perpendicular to the line 5y + 4x = 35 ​
Shkiper50 [21]

Step-by-step explanation:

<h2><em><u>concept :</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10</u></em></h2><h2 /><h2><em><u>concept :When two lines are perpendicular, then the product of their slopes is equivalent to -1.Equation of line in the form y = mx + c have m as slope of line and c as y-intercept.Solution:Given equations of lines are4y = 5x-10or, y = (5/4)x(5/2).</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>1</em><em>)</em></h2><h2 /><h2><em><u>5y + 4x = 35</u></em></h2><h2 /><h2><em><u>5y + 4x = 35ory = (-4/5)x + 7.</u></em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>(</em><em>2</em><em>)</em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1</u></em></h2><h2 /><h2><em><u>Let m and n be the slope of equations i and ii, respectively.Here, m = 5/4n= -4/5therefore, mx n = -1Hence, the lines are perpendicular.</u></em></h2>
8 0
3 years ago
Solve This Using The Quadratic Formula: 3x^2+9x-6=0
aivan3 [116]
<h3>Therefore either  x = \frac{-3+\sqrt{17}}{2}    or,  x = \frac{-3-\sqrt{17}}{2}</h3>

Step-by-step explanation:

3x^2 +9x -6=0                                x = \frac{-b\pm\sqrt{b^2- 4ac} }{2a}

\Leftrightarrow x = \frac{-9\pm\sqrt{9^2- 4.3(-6)} }{2.3}                     here a = 3 ,b = 9 and c= -6

\Leftrightarrow x = \frac{-9\pm\sqrt{153} }{6}

\Leftrightarrow x = \frac{3(-3\pm\sqrt{17}) }{6}

\Leftrightarrow x = \frac{-3\pm\sqrt{17}}{2}

Therefore either  x = \frac{-3+\sqrt{17}}{2}    or,  x = \frac{-3-\sqrt{17}}{2}

4 0
3 years ago
How to answer this question
Sonbull [250]
Look at picture for details

6 0
3 years ago
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