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Paladinen [302]
3 years ago
12

2.

Mathematics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

2. Ratio of 3/4

4. Ratio of 1/2

Step-by-step explanation:

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Find the vertex and axis of symmetry of each quadratic equation.
icang [17]

Answer: Vertex (-2,-15) and therefore the axis of sym will be -2

Step-by-step explanation: Using -b/2a you can deduce that 4x^2 is a, 16x is b and 1 is c. So -b/2a = -16/8 = -2. Then you plug -2 for y and yu should get -15. Then x will be your axis of symetry to x=-2

4 0
1 year ago
2*8*5=2*5*8 identify the property that is shown
ira [324]
The property shown is commutative multiplication.  This property says that the product of two factors is not affected by the order in which they're multiplied.  In this case, the product remains the same i.e.,  2*8*5 = 80 while 2*5*8 <em>also </em>equals 80.
7 0
3 years ago
Read 2 more answers
The table below shows points that are on the graph of the function h(x). 0 1 2 3 4 5 h(2) 2 5 8 11 14 17 Select the point that i
sukhopar [10]

9514 1404 393

Answer:

  B.  (2, 0)

Step-by-step explanation:

We understand the points on the graph of h(x) to be ...

  (0, 2), (1, 5), (2, 8), (3, 11), (4, 14), (5, 17)

The points on the graph of the inverse function will have the x- and y-coordinates reversed:

  (2, 0), (5, 1), (8, 2), (11, 3), (14, 4), (17, 5)

The one that appears among the answer choices is ...

  (2, 0)

7 0
2 years ago
100 POINTS AND BRAINLIEST
zysi [14]

Answer:

8 square units and \frac{40}{3} square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor \frac{1}{3}, then

\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.

2. Triangles ABC and DBE are similar with scale factor \frac{2}{3}, then

\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.

3. Thus, the area of the quadrilateral DFGE is

A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.

and the area of the quadrilateral ADEC is

A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.

4 0
3 years ago
Read 2 more answers
Simplify the expression below. 14a8y3 − 7a4y5 + 28a12y2 7a4y
olchik [2.2K]

Answer:

14a^{8}y^{3} - 7a^{4}y^{5} + 4a^{8} y

Step-by-step explanation:

14a^{8}y^{3} - 7a^{4}y^{5} + \frac{28a^{12}y^{2}}{7a^{4} y}

14a^{8}y^{3} - 7a^{4}y^{5} + 4a^{8} y

5 0
3 years ago
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