All categories

2 years ago

Which kind of function best models the data in the table? Graph the data and write an equation to model the data.

Mathematics

2 answers:

mestny [16]2 years ago

8 0

linear; y = x – 2 is answer! pls mark me as brainliest i need only 2 more to rank up! Hope I Helped

Firdavs [7]2 years ago

3 0

Answer: the answer is x-2

Step-by-step explanation:

You might be interested in

What is 2 1/2 - 1 1/8

Lady bird [3.3K]

Answer:

<h2>Convert the <em>mixed numbers</em> to </h2><h2><em>improper fractions</em>, </h2><h2>then find the <em>LCD</em> and combine.</h2>

<u />

<u>Exact Form:</u>

11/8

<u>Decimal Form:</u>

1.375

<u>Mixed Number Form:</u>

1 3/8

4 0

2 years ago

Simplify √ 20 x 8 y 9 to the form a √ b .

Contact [7]

Answer:

Juanita dissolves 46 g of MgBr2 (molar mass: 184.11 g/mol) in 0.5 kg of distilled water. What is the molality of the solution?

6 0

2 years ago

Points A, B, C, and D are collinear and positioned in that order. Find the length indicated 7) Find BC if BC = 2x + 4, AD = 28,

belka [17]

Answer:

7) BC = 10

8) BD = 20

Step-by-step explanation:

7) The segment addition theorem tells you ...

AB +BC +CD = AD

(3x+2) +(2x+4) +(3x-2) = 28

8x +4 = 28 . . . . collect terms

8x = 24 . . . . . . . subtract 4

x = 3 . . . . . . . . . divide by 8

BC = 2x+4 = 2(3) +4

BC = 10

8) AB +BD = AC +CD

(2x -14) +(-7 +3x) = (2x -3) +(9)

5x -21 = 2x +6

3x = 27

x = 9

BD = -7 +3x = -7 +3(9)

BD = 20

7 0

3 years ago

How to evaluate the expression

jasenka [17]

Answer:

Step-by-step explanation:

$4^{5}$ can be expressed as $2^{(2)(5)}$ = $2^{10}$

Similarly $4^{8}$ can be expressed as $2^{(2)(8)}$ = $2^{16}$

Numerator becomes:

$2^{10}$ · $-2^{9}$ = $-2^{19}$

Denominator becomes:

$2^{16}$ · $-2^{3}$ = $-2^{19}$

Since numerator = Denominator,

Answer = 1

Edit reason: typo

4 0

2 years ago

What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE

Nutka1998 [239]

Answer:

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),$ $\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

Step-by-step explanation:

Let points D, E and F have coordinates $(x_D,y_D),\ (x_E,y_E)$ and $(x_F,y_F).$

1. Midpoint M of segment DF has coordinates

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).$

2. Midpoint N of segment EF has coordinates

$\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0

2 years ago