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mel-nik [20]
3 years ago
6

How to solve a one variable equation

Mathematics
1 answer:
scZoUnD [109]3 years ago
5 0
You would isolate the variable for ex

2x+3=9
2x=9-3
2x=6 
x=3
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Hi can someone please help me with this. It urgent and time tysm.
navik [9.2K]
2x=3x-29

-x=-29
Divide by negative one

x=29.
3 0
3 years ago
Write the point-slope form AND slope-intercept form equations for the line that passes through the points (2, -4) and (3, -7)
never [62]

Answer:

answer is picture and shown

6 0
1 year ago
How do u solve 4(3x-1)=9-x
Nastasia [14]

Answer:

x = 1

Step-by-step explanation:

Well the first step is to apply the distributive property:

4(3x - 1) is equal to (12x - 4). You DISTRIBUTE a '4' to what is inside the parentheses.

And btw, to make it easier, you can make 9 - x so that x is first. For example, (-x + 9). They're both the same thing, just written differently.

Your new equation is 12x - 4 = -x + 9. You want to now solve the equation.

Add the (-x) to both sides. It cancels out on the right side and you add it to 12x on the left side.

[If there's no number in front of a variable, you can always just put 1 in order to make it easier]

12x + 1x = 13x. Your new equation is 13x - 4 = 9. This should look very familiar. You simply add 4 to both sides. 9 + 4 = 13

Finally, 13x = 13. Divide 13 ÷ 13 to get 1.

x = 1

5 0
3 years ago
The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged
Temka [501]

Answer:

a) 0.283  or 28.3%

b) 0.130 or 13%

c) 0.4 or 40%

d) 30.6 mm

Step-by-step explanation:

z-score of a single left atrial diameter value of healthy children can be calculated as:

z=\frac{X-M}{s} where

  • X is the left atrial diameter value we are looking for its z-score
  • M is the mean left atrial diameter of healthy children (26.7 mm)
  • s is the standard deviation (4.7 mm)

Then

a) proportion of healthy children who have left atrial diameters less than 24 mm

=P(z<z*) where z* is the z-score of 24 mm

z*=\frac{24-26.7}{4.7} ≈ −0.574

And P(z<−0.574)=0.283

b) proportion of healthy children who have left atrial diameters greater than 32 mm

= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm

z*=\frac{32-26.7}{4.7} ≈ 1.128

1-P(z<1.128)=0.8703=0.130

c) proportion of healthy children have left atrial diameters between 25 and 30 mm

=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm

z(30)=\frac{30-26.7}{4.7} ≈ 0.702

z(25)=\frac{25-26.7}{4.7} ≈ −0.362

P(z<0.702)=0.7587

P(z<−0.362)=0.3587

Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4

d) to find the value for which only about 20% have a larger left atrial diameter, we assume

P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.

Then P(z<z*)=0.8 and z*=0.84. That is

0.84=\frac{X-26.7}{4.7}  solving this equation for X we get X=30.648

5 0
3 years ago
Curt and Ian both ran a mile. Curt time was 8/9 of Ian time. Who ran faster?
Leona [35]

Curt because if his time was 8/9 of Ian's time then that means he ran the mile in a fraction of Ian's time, so he finished the mile quicker. Curt ran faster.

5 0
3 years ago
Read 2 more answers
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