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iragen [17]
3 years ago
11

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at the

ir math club meeting. They paid their guest speaker $1−A−2−$1_A_2_. What is the missing digit AA of this 33-digit number?
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
6 0
The correct answer for the question that is being presented above is this one: "<span>The missing digit is 3."</span>

The missing digit is represented by A  ⇒  $1A2, where A represents a whole number from 0 to 9.

Try substituting any number from 0 to 9 for A (the middle digit or tens place) , then divide by 11 members.

<span>Option 2:  A = 3  ⇒  $1 </span><span>3 </span>2
$ 132 ÷ 11 = $12

The only one-digit that gives a whole number quotient when it takes the place of tens digit in $1 __2  is<span> 3.</span><span>   Because $ 132 ÷ 11 = $ 12. (Option 2)</span>

The missing digit is 3.
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tomas wrote an equation 4/5x-8=. Finish the equation so that the equation will have no solution. Explain how you know.
Pepsi [2]

Answer:

Right part: \dfrac{4}{5}x+a, where a\neq -8.

Step-by-step explanation:

Linear equation have no solution if it is impossible for the equation to be true no matter what value we assign to the variable x.

The left part of the equation Tomas wrote is

\dfrac{4}{5}x-8.

The right part of the equation should be of the form

\dfrac{4}{5}x+a,

where a\neq -8.

In this case, the equation will take look

\dfrac{4}{5}x-8=\dfrac{4}{5}x+a,\\ \\-8=a.

Since a\neq -8, this equality is always false.

Remark: 1) If a=8, the equation \dfrac{4}{5}x-8=\dfrac{4}{5}x+a is equivalent to the equation

\dfrac{4}{5}x-8=\dfrac{4}{5}x-8,\\ \\0=0

and the lest equation has infinitely many solutions.

2) When coefficient at x differs from \dfrac{4}{5}, the equation has unique solution.

6 0
3 years ago
100 POINTS
Shkiper50 [21]

Answer:

  6√3 ±3 ≈ {7.392, 13.392}

Step-by-step explanation:

The length of AB is the long side of a right triangle with hypotenuse CD and short side (AC -BD). The desired radius values will be half the length of EF, with AE added or subtracted.

__

<h3>length of AB</h3>

Radii AC and BD are perpendicular to the points of tangency at A and B. They differ in length by AC -BD = 12 -9 = 3 units.

A right triangle can be drawn as in the attached figure, where it is shaded and labeled with vertices A, B, C. Its long leg (AB in the attachment) is the long leg of the right triangle with hypotenuse 21 and short leg 3. The length of that leg is found from the Pythagorean theorem to be ...

  AB = √(21² -3²) = √432 = 12√3

<h3>tangent circle radii</h3>

This is the same as the distance EF. Half this length, 6√3, is the distance from the midpoint of EF to E or F. The radii of the tangent circles to circles E and F will be (EF/2 ±3). Those values are ...

  6√3 ±3 ≈ {7.392, 13.392}

5 0
2 years ago
Read 2 more answers
I need The answers now hellllp please ​
lisabon 2012 [21]

Answer:

Step-by-step explanation:

6A = 60°

6B = 34°

6 0
3 years ago
Read 2 more answers
What are the solutions to the system of equations graphed below?
Morgarella [4.7K]

Answer:

looking at the straight line, it touches the axis at (-2,0) and (6,0)

Step-by-step explanation:

6 0
3 years ago
Can someone help me with A, B and C
MariettaO [177]
Find two factors of the last number that add up to the middle number.
a) two factors of 16 that add to 10: 8 and 2. Because 8x2=16 and 8+2=10. then do this:
(x+8)(x+2)

b) (x+4)(x+3)

c) (x+12)(x+1)
3 0
4 years ago
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