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kirza4 [7]
3 years ago
5

The sum of two numbers is 179. Their difference is 133. Find both numbers.

Mathematics
2 answers:
Marianna [84]3 years ago
8 0

Answer:

156,23

Step-by-step explanation:

let the numbers be x and y and x>y

x+y=179

x-y=133

add

2x=312

x=156

156+y=179

y=179-156=23

Savatey [412]3 years ago
7 0

Answer:

156 and 23

Step-by-step explanation:

  • We can use system of equations to solve this problem.

Step 1: Set up the system of equations.

  • \left \{ {{x+y=179} \atop {x-y=133}} \right.

Step 2: Solve x + y = 179 for x.

  • x + y = 179
  • x + y - y = 179 - y (Subtract y from both sides)
  • x = -y + 179

Step 3: Substitute -y + 179 for x in x - y = 133.

  • x-y=133
  • -y+179-y=133
  • (-y-y)+(179)=133 (Combine like terms)
  • -2y + 179 = 133
  • -2y + 179 - 179 = 133 - 179 (Subtract 179 from both sides)
  • -2y = -46
  • \frac{-2y}{-2} = \frac{-46}{-2} (Divide both sides by -2)
  • y = 23

Step 4: Substitute 23 for y in x = -y + 179

  • x = -23 + 179
  • x = 156

Therefore, the numbers are 156 and 23.

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barxatty [35]

Answer:

assuming its an annual interest

Okay so 6 percent interest, the bank is paying you.
So with this it’s 6 percent of 1500 and add it to 1500.

You can always find 6 percent of 1500 and then add but here’s a short cut.
Your principle (beginning) balance is 1500.
That’s already 100 percent since thats yoru original value.
You then get added 6 percent interest.
We are jsut adding 6 percent to 100 percent so 106 percent.
Now we solve normally and you’d get the answer faster.

106 percent is 106/100 or 1 3/5 or 1.06

now we multiply
1500 * 1.06 = 1590

Your final balance would be 1590 after the 6 percent interest is added.

5 0
2 years ago
Point B has coordinates ​(​1,2​). The​ x-coordinate of point A is -8. The distance between point A and point B is 15 units. What
Xelga [282]

Answer:

The possible coordinates of point A are A_{1} (x,y) = (-8, 14) and A_{2} (x,y) = (-8, -10), respectively.

Step-by-step explanation:

From Analytical Geometry, we have the Equation of the Distance of a Line Segment between two points:

l_{AB} = \sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2}} (1)

Where:

l_{AB} - Length of the line segment AB.

x_{A}, x_{B} - x-coordinates of points A and B.

y_{A}, y_{B} - y-coordinates of points A and B.

If we know that l_{AB} = 15, x_{A} = -8, x_{B} = 1 and y_{B} = 2, then the possible coordinates of point A is:

\sqrt{(1+8)^{2}+(2-y_{A})^{2}} = 15

81 + (2-y_{A})^{2} = 225

(2-y_{A})^{2} = 144

2-y_{A} = \pm 12

There are two possible solutions:

1) 2-y_{A} = -12

y_{A} = 14

2) 2 - y_{A} = 12

y_{A} = -10

The possible coordinates of point A are A_{1} (x,y) = (-8, 14) and A_{2} (x,y) = (-8, -10), respectively.

8 0
3 years ago
In a unit circle, ø = 270º. Identify the terminal point and sin ø.
larisa [96]

Answer:

Terminal point (0, -1); sin Ø = -1 ⇒ A

Step-by-step explanation:

In the unit circle, Ф is the angle between the terminal side and the positive part of the x-xis

  • The terminal point on the positive part of the x-axis is (1, 0),which means Ф = 0° or 360° and cosФ = 1, sinФ = 0
  • The terminal point on the positive part of the y-axis is (0, 1),which means Ф = 90° and cosФ = 0, sinФ = 1
  • The terminal point on the negative part of the x-axis is (-1, 0),which means Ф = 180° and cosФ = -1, sinФ = 0
  • The terminal point on the negative part of the y-axis is (0, -1),which means Ф = 270° and cosФ = 0, sinФ = -1

In a unit circle

∵ Ф = 270°

→ By using the 4th rule above

∴ The terminal point is (0, -1)

∴ sinФ = -1

∴ Terminal point (0, -1); sin Ø = -1

4 0
3 years ago
Read 2 more answers
Determine whether (2, - 1) and (-4, 2) satisfy the inequality 2x – 3y > 4.
Xelga [282]

Answer:

emngtehrge

Step-by-step explanation:

egrbrrg

7 0
3 years ago
which is the rationalized form of the expression the square root of x divided by the square root of x +the square root of 7
Ilya [14]

Answer:

I assume your goal is to rationalize the denominator -

So you need to multiply the denominator by its conjugate.

So we have:

\frac{\sqrt{x} }{\sqrt{x}-\sqrt{7} } *\frac{\sqrt{x} +\sqrt{7} }{\sqrt{x} +\sqrt{7}}

->

\frac{x-\sqrt{7}\sqrt{x} }{x-7}

gg

8 0
3 years ago
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