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frez [133]
3 years ago
12

9° TERMINO DE 7.10.13​

Mathematics
2 answers:
andrew-mc [135]3 years ago
3 0

Answer:

djkdjdjfifkflkfkfuiririrkrkkr kfkrkrkrlrlrlrlrlrlfkkrkfllrkrkrkkrkfkkfkkrkrkrkkrkrkkr krkrkkkkrkfkkkkfkrkkfkfkfkkf ir rkkfikriffiriifi

ycow [4]3 years ago
3 0

Answer:

Evaluate

18 ER TM×2,130

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I was stuck on the same thing in my class test. I ended up failing but if I get the answers to it I’ll totally send them to you
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The quadratic function h(t) = -16t^2 + 120 represents the height of 120 feet. what is the height of the object after 1.5 seconds
Anettt [7]

Answer:

84 feet

Step-by-step explanation:

-16(1.5)^2+120=84

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3 years ago
Does the graph represent a function?
Genrish500 [490]

Answer:

No it's not a function. Functions can't have the same x value.

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3 years ago
Read 2 more answers
34
Vadim26 [7]

Answer:

They are similar by the definition of similarity in terms of a dilation

Step-by-step explanation:

The given vertices of triangle ΔABC are;

A(1, 5), B(3, 9), and C(5, 3)

The vertices of triangle ΔDEF are;

D(-3, 3), E(-2, 5), and F(-1, 2)

Therefore, we get;

The length of segment, \overline{AB} = √((9 - 5)² + (3 - 1)²) = 2·√5

The length of segment, \overline{BC} = √((9 - 3)² + (3 - 5)²) = 2·√10

The length of segment, \overline{AC} = √((5 - 3)² + (1 - 5)²) = 2·√5

The length of segment, \overline{DE} = √((5 - 3)² + (-2 - (-3))²) = √5

The length of segment, \overline{EF} = √((2 - 5)² + (-1 - (-2))²) = √10

The length of segment, \overline{DF} = √((2 - 3)² + (-1 - (-3))²) = √5

∴ \overline{AB}/\overline{DE} = 2·√5/(√5) = 2

\overline{BC}/\overline{EF} = 2·√10/(√10) = 2

\overline{AC}/\overline{DF} = 2·√5/(√5) = 2

The ratio of their corresponding sides are equal and therefore;

ΔABC and ΔDEF are similar by the definition of similarity in terms of dilation.

4 0
3 years ago
Math 8: Linear Functions, Part 2
Studentka2010 [4]

Answer:

\boxed{\text{1. y + 5 = -4(x - 3); \qquad 2. y - 8 = x + 1}}

Step-by-step explanation:

Question 1

The point-slope formula for a straight line is

y – y₁ = m(x – x₁)

x₁ = 3; y₁ = -5; m = -4  

Substitute the values

\boxed{\textbf{y + 5 = -4(x - 3)}}

The diagram shows the graph of equation 1 (red) with slope -4 passing through (3,-5).

Question 2

x₁ = -1; y₁ = 8; m = 1  

Substitute the values

\boxed{\textbf{y - 8 = x + 1}}

The diagram shows the graph of equation 2 (green) with slope 1 passing through (-1,8).

3 0
4 years ago
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