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professor190 [17]
3 years ago
8

Which of the following word problems can be solved using the equation 10 x - 15=60

Mathematics
1 answer:
Alja [10]3 years ago
8 0

Answer:

Kit buys

Step-by-step explanation:

im right

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Wich points are 4 units apart? (Image provided)
Andrew [12]

Answer:

D, B

Step-by-step explanation:

D is 4 units above B

6 0
3 years ago
Prove that the diagonals of a parallelogram bisect each other​
Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof :\begin{array}{ |c| c |  c |  } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In  \: \triangle ^{s}  \:AOB \: and \: COD  } \\  \sf{(i)}&  \sf{ \angle \: OAB =  \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\  \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\  \sf{(iii)} &\sf{ \angle \: OBA=  \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\  \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}.          Proved ✔

♕ And we're done! Hurrayyy! ;)

# STUDY HARD! So, Tomorrow you can answer people like this , " Dude , I just bought this expensive mobile phone but it is not that expensive for me" [ - Unknown ] :P

☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

\underbrace{ \overbrace  {\mathfrak{Carry \: On \: Learning}}} ☂

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

5 0
3 years ago
Can someone tell me if it’s correct?
mart [117]
Yes yes it is correct
8 0
3 years ago
Read 2 more answers
Solve the equation 5x+2-x=-4x
AVprozaik [17]
5x + 2 -x = -4x

4x +2 = -4x

2 = -4x -4x

2 = -8x

x = - 1/4


hope this helps
8 0
3 years ago
Read 2 more answers
How do you solve 58 and 1/3÷6 and 2/3
strojnjashka [21]

Answer:

35/4, or 8 3/4 as a mixed number

Step-by-step explanation:

First, change 58 1/3 into an improper fraction by multiplying the whole number and denominator, then adding the numerator. 58 x 3 equals 174, + 1 equals 175. So, 58 1/3 as an improper fraction is 175/3. Next, change 6 2/3 into an improper fraction. 6 times 3 equals 18, plus 2 equals 20. So, it's 20/3. So, here are your two fractions:

175/3 & 20/3

To divide fractions, I like to use a method called Keep Change Flip. Basically, you keep the first fractions the same, then change the sign. The division sign changes into a multiplication symbol. Now, your equation should look like this: 175/3 x 20/3. Next, flip the fraction from 20/3 into 3/20. This is what your equation should look like now: 175/3 x 3/20. Now you can multiply the fractions together. Before you do so, you can cross reduce to make it easier. What is a number that both 3 and 3 can be divided by? The correct answer is 3. 3/3= 1, so the equation is now 175/1 x 1/20. However, you can continue to cross reduce. You can also divide 175 and 20 by 5, so the equation changes into this: 35/1 x 1/20. Multiply numerator by numerator, denominator by denominator. So, the answer is 35/4, or 8 3/4 as a mixed number. Hope this helped!

5 0
3 years ago
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