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yaroslaw [1]
3 years ago
11

Enter Brackets To Make The Statement 10 + 2 x 3²– 2 = 254

Mathematics
1 answer:
muminat3 years ago
3 0

Consider expression (10+2\cdot 3)^2-2:

1) 2\cdot 3=6;

2) 10+2\cdot 3=10+6=16;

3) (10+2\cdot 3)^2=16^2=256;

4) (10+2\cdot 3)^2-2=256-2=254.

Answer: (10+2\cdot 3)^2-2=254.

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7 1/2÷(4 1/2 - 5 1/8)
s2008m [1.1K]

Answer:

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Step-by-step explanation:

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3 years ago
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Can someone write an equation for me ?
sammy [17]
The correct answer is f=(1/3)m
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Which graph shows the rotation of the shape above, 90° counterclockwise about the origin?​
lara31 [8.8K]

Answer:

W

Step-by-step explanation:

the rule of a 90⁰ counterclockwise rotation is (x,y) -> (-y,x)

For example, if you rotated a point at (3,2) by 90⁰ counterclockwise. You would change it to (-2,3).

5 0
3 years ago
What is an equation in slope-intercept form of the line that passes through the points (−2, −2) and (1, 7)? A.y = 4x + 3 B.y − 7
antoniya [11.8K]

Answer:

C.y = 3x + 4

Step-by-step explanation:

We have two points, so we can find the slope

m = (y2-y1)/ (x2-x1)

   = (7--2)/ (1--2)

   = (7+2)/(1+2)

    = 9/3

    =3

The slope is 3

We can find the point slope form of the line

y-y1 = m(x-x1)

y-7 = 3(x-1)

Distribute

y-7 =3x-3

Add 7 to each side

y-7+7 = 3x-3+7

y = 3x+4

This is in slope intercept form (y=mx+b)

8 0
3 years ago
I’m need this worked out step by step by tonight
Pie

9514 1404 393

Answer:

  -3 ≤ x ≤ 19/3

Step-by-step explanation:

This inequality can be resolved to a compound inequality:

  -7 ≤ (3x -5)/2 ≤ 7

Multiply all parts by 2.

  -14 ≤ 3x -5 ≤ 14

Add 5 to all parts.

  -9 ≤ 3x ≤ 19

Divide all parts by 3.

  -3 ≤ x ≤ 19/3

_____

<em>Additional comment</em>

If you subtract 7 from both sides of the given inequality, it becomes ...

  |(3x -5)/2| -7 ≤ 0

Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.

__

For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...

  |f(x)| < (some number) . . . . or ≤

If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:

  |f(x)| > g(x)   ⇒   f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0

Any solutions to these inequalities must respect their domains.

8 0
3 years ago
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