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VladimirAG [237]
2 years ago
7

Isolate the variable on the following problem. show your work.11z= 4​

Mathematics
2 answers:
azamat2 years ago
5 0
(1/11)(11z)= 4(1/11)
Z= 4/11 is the answer in fraction form
rjkz [21]2 years ago
4 0
Divide by 11 to make z by itself
z = 4/11
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This set of points is on the graph of a function.
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(-1,1) and (0,0) are probably it
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3 years ago
What's the simplist way to simplify 13/78
Andrews [41]
13/78 simplified is 1/6
4 0
3 years ago
Read 2 more answers
What is the answer to y=(2x+5)(x-2) in standard form
docker41 [41]

Answer:

i think it is Rewrite in Standard Form y=2x+5. y=2x+5 y = 2 x + 5.

Step-by-step explanation:

5 0
3 years ago
1. The sum of a two-digits number is 13. The tens digit is 8 less than twice the units digit. What is the number?
alexgriva [62]

Answer:

<u>The number is 67</u>

Step-by-step explanation:

<u>Equations</u>

Let's consider the number 83. The tens digit is 8 and the unit digit is 3. Note the tens digit's addition to the number is 80, and the unit's addition is 3. This means the tens digit adds 10 times its value, that is, 83 = 8*10 + 3.

Now, let's consider the number ab, where a is the tens digit, and b is the unit digit. It follows that

Number=10*a+b

The question gives us two conditions:

1) The sum of a two-digits number is 13.

2) The tens digit is 8 less than twice the units digit.

The first condition can be expressed as:

a + b = 13                     [1]

And the second condition can be written as:

a = 2b-8                      [2]

Replacing [2] into [1], we have:

2b-8 + b = 13

Operating:

3b = 13 + 8

3b = 21

Solving for b:

b = 21 / 3 = 7

Substituting into [2]:

a = 2*(7) - 8 = 6

Thus, the number is 67

3 0
3 years ago
Another fair dice is rolled 300 times. How many times would you expect it to land with a multiple of three on top? I need help t
ira [324]

Answer:

100 times

Step-by-step explanation:

multiples of 3 on a number cube are 3 and 6

probability of rolling a '3' or '6' is 1/6 + 1/6 = 2/6 or 1/3

1/3 x 300 = 100

7 0
3 years ago
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