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nalin [4]
3 years ago
7

Need your help! Question in picture :)

Mathematics
2 answers:
Wewaii [24]3 years ago
6 0

56666666666g c5vtfyg

zhenek [66]3 years ago
5 0

Answer:

the number is 70

Step-by-step explanation

80% × ? = 56

? =

56 ÷ 80% =

56 ÷ (80 ÷ 100) =

(100 × 56) ÷ 80 =

5,600 ÷ 80 =

70;

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cestrela7 [59]

Answer:

8

Step-by-step explanation:

4 0
3 years ago
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Clinton and Katie order some food at a food truck. At this food truck Clinton buys 6 apple pies and 2 juices for $39. Katie buys
Tamiku [17]

________________________________

<h2>Pie: </h2><h3>up = m ÷ ti</h3><h3>= $57 ÷ 8 Pies</h3><h3>= 7.125</h3><h3>= $39 ÷ 7.125</h3><h3>= ~ $5.47368</h3><h3>= $6</h3>

<h2>Juice: </h2><h3>up = m ÷ ti</h3><h3>= $57 ÷ 6 Juices</h3><h3>= 9.5</h3><h3>= $18 ÷ 9.5</h3><h3>= ~ $1.89473</h3><h3>= $1.50</h3>

<h2>The Price Of An Apple Pie Is $6 Each & The Price Of A Juice Is $1.50 Each.</h2>

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8 0
3 years ago
Look at the attachment<br>Simplified it too <br>pls take it seriously ​
Phantasy [73]

\\ \rm\rightarrowtail \dfrac{sin(180+A)+2cos(180+A).cos(A-180)}{2cos^2(360+A)-cos(-A)}

\\ \rm\rightarrowtail \dfrac{-sinA+2(cos^2A-sin^2A)}{2cos^2A-cosA}

\\ \rm\rightarrowtail \dfrac{-sinA+2cos^2A-2sin^2A}{cosA(2cosA-1)}

\\ \rm\rightarrowtail \dfrac{-sinA+2-2sin^2A-2sin^2A}{cosA(2cosA-1)}

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5 0
1 year ago
PLEASE HELP THE QUESTION IS BELOW!!
irina [24]

Answer:

Step-by-step explanation:

B=3x-2

C=12x+2

D=B=3x-2 (Vertically opposite angles)

A=?

Now; we know, A+B+C+D=360 deg.

=>A=360-(B+C+D)

  =360-(3x-2+12x+2+3x-2)

  =360-(18x-2) ----[1]

B+C=180 deg. (linear pair)

=> 3x-2+12x+2=180

=> 15x=180

=> x=180/15

   =12 ---[2]

subsitute [2] in [1];

=> A= 360-(18x-2)

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So, the correct answer is (D).

Hope the answer is useful.  

8 0
3 years ago
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Can we obtain a diagonal matrix by multiplying two non-diagonal matrices? give an example
polet [3.4K]
Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.

Consider the matrix multiplication below

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]   \left[\begin{array}{cc}e&f\\g&h\end{array}\right] =  \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]

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Consider the following sets of values

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