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2 years ago

4.50x +5.50x – 7.50x

Mathematics

1 answer:

Wittaler [7]2 years ago

3 0

Hey there!

4.50x + 5.50x - 7.50x

= 4.5x + 5.5x - 7.5x

= 10x - 7.5x

COMBINE the LIKE TERMS

= (10x - 7.5x)

= 10x - 7.5x

= 2.5x

≈ 2.50x

Therefore, your answer is: 2.50x

Good luck on your assignment & enjoy your day!

~Amphitrite1040:)

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On a coordinate plane, 2 quadrilaterals are shown. The first figure has points A (negative 2, 1), B (negative 4, 1), C (negative

Afina-wow [57]

The rule of reflection is ry-axis (x , y) → (-x , y) ⇒ 2nd answer

Step-by-step explanation:

Let us revise some transformation

If point (x , y) reflected across the x-axis , then its image is (x , -y) , the rule of reflection is rx-axis (x , y) → (x , -y)
If point (x , y) reflected across the y-axis , then its image is (-x , y) , the rule of reflection is ry-axis (x , y) → (-x , y)

∵ The first quadrilateral has vertices

A (-2 , 1) , B (-4 , 1) , C (-4 , 5) , D (-2 , 4)

∵ The second quadrilateral has vertices

A' (2 , 1) , B' (4 , 1) , C' (4 , 5) , D' (2 , 4)

- The x-coordinates of point A' , B' , C' , D' have the same values

and opposite signs of the x-coordinates of points A , B , C , D

∴ A' , B' , C' , D' are the images of A , B , C , D after reflection

across the y-axis

∴ The rule of reflection is ry-axis (x , y) → (-x , y)

The rule of reflection is ry-axis (x , y) → (-x , y)

Learn more:

You can learn more about reflection in brainly.com/question/5017530

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3 years ago

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What is the simplified form of V10,000 ,

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2 years ago

What are the domain and range of the function f(x) = 2^x+1?

scoundrel [369]

Answer:

Domain: (-∞, ∞)

Range: (0,∞)

Step-by-step explanation:

Exponential functions are curves which approach a horizontal asymptote usually at y=0 or the x-axis unless a value has been added to it. If it has, the curve shifts. This function has addition on the exponent but not to the whole function so it does not change the asymptote. Its y - values remain between 0 and ∞. This is the range, the set of y values.

However, the range of exponentials can change based on the leading coefficient. If it is negative the graph flips upside down and its range goes to -∞. This doesn't have it either.

The addition to 1 on the exponent shifts the function to the left but doesn't change the range.

In exponential functions, the x values are usually not affected and all are included in the function. Even though it shifts, the domain doesn't change either. Its domain is (-∞, ∞).

Domain: (-∞, ∞)

Range: (0,∞)

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3 years ago

All 3 questions 3 pictures for find the missing angles, please with reasoning why? 100 points

Katyanochek1 [597]

Answer:

Problem 1)

$m\angle DEG = 38^\circ$

Problem 2)

$m\angle x =140^\circ \text{ and } m\angle y = 49^\circ$

Problem 3)

$m\angle EDG = 65^\circ$

Step-by-step explanation:

Problem 1: ∠DEG

From the diagram, we know that ∠DFG intercepts Arc DG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

$\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=m\angle DFG$

We know that m∠DFG = 38°. So:

$\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=38\Rightarrow m\stackrel{\frown}{DG}=76^\circ$

∠DEG also intercepts Arc DG. Hence:

$\displaystyle m\angle DEG=\frac{1}{2}m\stackrel{\frown}{DG}$

We know that Arc DG measures 76°. Hence:

$\displaystyle m\angle DEG =\frac{1}{2}\left(76\right)=38^\circ$

Alternate Explanation:

Since ∠DEG and ∠DFG intercept the same arc, ∠DEG ≅ DFG. So, m∠DEG = m∠DFG = 38°.

Problem 2: Circle with Centre O

(Let the bottom left corner be A, upper be B, and right be C.)

∠ACB intercepts Arc AC.

Inscribed angles have half the measure of its intercepted arc. Therefore:

$\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=m \angle ACB$

Since m∠ACB = 70°:

$\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=70\Rightarrow m\stackrel{\frown}{AC}=140^\circ$

∠x is a central angle and also intercepts Arc AC.

The measure of a central angle is equal to its intercepted arc. Thus:

$\displaystyle m\angle x =m\stackrel{\frown}{AC}=140^\circ$

The sum of the interior angles of a polygon is given by the formula:

$(n-2)\cdot 180^\circ$

Where n is the number of sides.

Since the inscribed figure is a four-sided polygon, its interior angles must total:

$(4-2)\cdot 180^\circ =360^\circ$

Therefore:

$21+70+y+(360-x)=360$

Substitute and solve for y:

$91+y+(360-140)=360\Rightarrow y +311=360\Rightarrow m\angle y = 49^\circ$

Problem 3: ∠EDG

∠EFG intercepts Arc EDG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

$\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG}=m\angle EFG$

Since m∠EFG = 115°:

$\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG} = 115\Rightarrow m\stackrel{\frown}{EDG} = 230^\circ$

A full circle measures 360°. Hence:

$m\stackrel{\frown}{EDG}+m\stackrel{\frown}{EFG}=360^\circ$

Since we know that Arc EDG measures 230°:

$230^\circ + m\stackrel{\frown}{EFG}=360$

Solve for Arc EFG:

$m\stackrel{\frown}{EFG}=130^\circ$

∠EDG intercepts Arc EFG.

Inscribed angles have half the measure of its intercepted arc. Therefore:

$\displaystyle m\angle EDG = \frac{1}{2}m\stackrel{\frown}{EFG}$

Since we know that Arc EFG measures 130°:

$\displaystyle m\angle EDG = \frac{1}{2}(130)=65^\circ$

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3 years ago

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Sauron [17]

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We see that the geese is 280, and 140 Godwits. In total 420.

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3 years ago

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