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4 years ago

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A tax cut aggregate demand and . A. decreases; shifts the AD curve leftward B. decreases; shifts the AD curve r

ightward C. does not change; does not shift the AD curve D. increases; shifts the AD curve leftward E. increases; shifts the AD curve rightward

1 answer:

aksik [14]4 years ago

5 0

Answer:

E. increases; shifts the AD curve rightward

Step-by-step explanation:

Tax cut means lower tax income tax rates, that mean individuals will have more of their income, that is more money to spend leading to increase in aggregate demand. This will lead to a rise (increase) in the aggregate demand curve which implies rightward movement.

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Help me Plz I’m stuck

spin [16.1K]

Answer:

Use all the given information:

<em>Price of a slice in $ -</em><em> x</em>

<em>Number of slices- </em><em>3</em>

<em>Total cost of slices-</em><em> 3x</em>

<em />

<em>Price of chopped cheese in $- </em><em> 2+x</em>

<em>Number of chopped cheeses-</em><em> 4</em>

<em>Total cost of chopped cheeses-</em><em> 8+4x</em>

<em />

<em>Total cost- </em><em>7x+8</em>

<em />

Step-by-step explanation:

We are told that the price of a slice is x and that Dashaun bought 3 slices.

So to find the total amount paid for slices, multiply the price by the number of slices:

$3*x=3x$

We are told that the price of chopped cheese is $2 more than a slice. Use y as the cost for chopped cheese:

$y=2+x$

Dashaun bought 4 chopped cheeses.

To find the total cost of chopped cheeses bought, multiply the cost per cheese by the amount of cheese purchased:

$4y$

Insert the value of y:

$4(2+x)$

Use the distributive property:

$8+4x$

This is the total cost for the chopped cheeses.

To find the total of lunch, add the prices for the slices and cheeses together:

$T=3x+8+4x\\\\T=7x+8$

The total cost for lunch is <em>7x+8</em>

:Done

4 0

3 years ago

Solve each quadratic equation by completing the square. 6. x2 + 2x = 8 7. x2 - 6x = 16 8. x2 - 18x = 19 9. x2 + 3x = 3 10. x2 +

Andrei [34K]

Lets get started :)

These questions have asked us to solve by completing the square.
How do we? I have attached a picture, which will explain

6. x² + 2x = 8
→ b is the coefficient of x, which is 2
→ We take half of 2 and square it. Then, we add it to either side

x² + 2x $+ (\frac{2}{2} )^2 = 8 + ( \frac{2}{2})^2$
x² + 2x + 1 = 8 + 1
( x + 1 )( x + 1 ) = 9
( x + 1 )² = 9
$\sqrt{( x + 1 )^2} = \sqrt{9}$
x + 1 = + 3 or x + 1 = - 3
x = 2 or x = - 4

7. x² - 6x = 16

→ We do the same thing we did in the previous question

x² - 6x + $(\frac{6}{2})^2 = 16 + (\frac{6}{2})^2$
x² - 6x + 9 = 16 + 9
(x - 3)² = 25
$\sqrt{(x-3)^2} = \sqrt{25}$
x - 3 = + 5 or x - 3 = - 5
x = 8 or x = - 2

8. x² - 18x = 19

x² - 18x + $( \frac{18}{2} )^2 = 19 + ( \frac{18}{2})^2$
x² - 18x + 81 = 19 + 81
( x - 9 )( x - 9 ) = 100
( x - 9 )² = 100
$\sqrt{(x-9)^2} = \sqrt{100}$
x - 9 = + 10 or x - 9 = -10
x = 19 or x = - 1

9. x² + 3x = 3

x² + 3x + $( \frac{3}{2} )^2 = 3 + (\frac{3}{2} )$
$x^2 + 3x + \frac{9}{4} = 3 + \frac{9}{4}$
$x^{2} + 3x + \frac{9}{4} = \frac{21}{4}$
$(x^2 + \frac{3}{2} ) ( x^2 + \frac{3}{2} ) = \frac{21}{4}$
$( x^2 + \frac{3}{2} )^2 = \frac{21}{4}$
$\sqrt{ x^2 + \frac{3}{2} } = \sqrt{ \frac{21}{4} }$
x + $\frac{3}{2}$ = + $\frac{ \sqrt{21} }{2}$ or x + $\frac{3}{2} = - \frac{ \sqrt{21} }{2}$
x = $\frac{-3+ \sqrt{21} }{2}$ or x = $\frac{-3 - \sqrt{21}}{2}$

7 0

3 years ago

Determine if segment AB is tangent to the circle.

Angelina_Jolie [31]

Answer: Segment AB is tangent to the circle

Step-by-step explanation:

3 sides are 16, 20, and 12

If angle a is 90 then ab is tangent

now check using Pythagorean theorem

since 12^2 + 16^2 =20^2, this means angle a is a right angle meaning segment AB is tangent to the circle

8 0

3 years ago

How do you divide 85 and 5?

AVprozaik [17]

Division is like sharing so 85 things that are shared between 5 people
85÷5=17 that is how much each person gets
so 17 is your answer

5 0

3 years ago

Read 2 more answers

Find the exact value, without a

alexira [117]

Answer:

$-2 -\sqrt{3}$

Step-by-step explanation:

<u><em>First consider numerator</em></u>

$sin \frac{\frac{7\pi}{6}}{2} = sin \frac{7\pi}{12}= sin (\frac{\pi}{4} + \frac{\pi}{3})$

Using the formula : sin (A + B) = sin A cos B + cos A sin B

$sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}$

$sin(\frac{\pi}{4} + \frac{\pi}{3}) = sin \frac{\pi}{4} \cdot cos \frac{\pi}{3} + cos \frac{\pi}{4} \cdot sin \frac{\pi}{3}$

$=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} + \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2} \\\\= \frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}\\\\=\frac{\sqrt{2} +\sqrt{6} }{4}$

<u><em>Second consider denominator</em></u>

$cos \frac{\frac{7\pi}{6}}{2} = cos \frac{7\pi}{12}= cos (\frac{\pi}{4} + \frac{\pi}{3})$

Using the formula : cos (A + B) = cos A cos B - sin A sin B

$sin \frac{\pi}{4} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{4} = \frac{\sqrt{2} }{2}\\\\sin \frac{\pi}{3} = \frac{\sqrt{2} }{2}, \ cos \frac{\pi}{3} = \frac{1}{2}$

$cos(\frac{\pi}{4} + \frac{\pi}{3}) = cos \frac{\pi}{4} \cdot cos \frac{\pi}{3} -sin \frac{\pi}{4} \cdot sin \frac{\pi}{3}$

$=\frac{\sqrt{2} }{2} \cdot \frac{1}{2} - \frac{\sqrt{2} }{2} \cdot \frac{\sqrt{3} }{2}\\\\=\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}\\\\= \frac{\sqrt{2} -\sqrt{6} }{4}$

Therefore,

$tan \frac{7\pi}{12} = \frac{sin \frac{7\pi}{12}}{cos\frac{7\pi}{12}}$

$= \frac{\frac{\sqrt{2} +\sqrt{6} }{4}} {\frac{\sqrt{2} -\sqrt{6} }{4} }\\\\=\frac{\sqrt{2} +\sqrt{6} }{4} \times \frac{4 }{\sqrt{2} -\sqrt{6}}\\\\=\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}}$

Either we can stop here or Rationalize the denominator:

$\frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} -\sqrt{6}} \times \frac{\sqrt{2} +\sqrt{6} }{\sqrt{2} +\sqrt{6}} = \frac{(\sqrt{2} +\sqrt{6})^{2} }{(\sqrt{2})^2 -(\sqrt{6})^2} = \frac{2 + 6 +2\sqrt{12} }{2-6} = \frac{8+2\sqrt{12} }{-4} = \frac{8+ 4\sqrt{3} }{-4} = -2-\sqrt{3}$

3 0

3 years ago