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mote1985 [20]
2 years ago
6

Place the numbers in ascending order:

Mathematics
1 answer:
goldenfox [79]2 years ago
6 0
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ The correct option would in fact be A. To help, you could always look at the negative decimals as negative whole numbers (take away the decimal point)

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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The inequality 6-2/3
mash [69]

16 over 3 is the answer u write the numerator above the denominator 6 over 1 minus 2 over 3 6 times 3 over 1 x3 minus 2 over 3 u get 18 minus 2 over 3 and subtract 18 and 2 u get 16 over 3

3 0
3 years ago
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Find the value of f(g(-2))
PSYCHO15rus [73]

Answer:

The correct ans is 3.

g(-2) = 2(-2) + 5

=-4 + 5 = 1

f(g(-2)) = 4 - (1)^2

= 4 - 1

=3

Step-by-step explanation:

4 0
2 years ago
F(x) = -4x - 3<br> Evaluate f(9) + f(5)=
RSB [31]

Evaluate f(9) + f(5)= 14f If I didn't answer Your Question fully than tell me please!

7 0
3 years ago
I need help please help me thanks please
Vitek1552 [10]

so we have a table of values, with x,y coordinates, so let's use any two of those points to get the slope of the table and use the point-slope form to get its equation

~\hspace{2.7em}\stackrel{\textit{let's use}}{\downarrow }\qquad \stackrel{\textit{and this}}{\downarrow }\\\begin{array}{|lr|r|r|r|r|}\cline{1-6}x&0&1&2&3&4\\y&-1&3&7&11&15\\\cline{1-6}\end{array}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad(\stackrel{x_2}{4}~,~\stackrel{y_2}{15})

\stackrel{slope}{m}\implies\cfrac{\stackrel{rise}{\stackrel{y_2}{15}-\stackrel{y1}{3}}}{\underset{run}{\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{12}{3}\implies 4\\\\\\% point-slope intercept\begin{array}{|c|ll}\cline{1-1}\textit{point-slope form}\\\cline{1-1}\\y-y_1=m(x-x_1)\\\\\cline{1-1}\end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{4}(x-\stackrel{x_1}{1})\\\\\\y-3=4x-4\implies y = 4x-1\implies \blacktriangleright  y = 4x+(-1)\blacktriangleleft

4 0
2 years ago
The table shows some values of a function of the form y = ax2 + bx + c.
natali 33 [55]

Answer:

Value of constant term c is (-4)

Step-by-step explanation:

The given table represents a function which is in the form of a quadratic equation,

y = ax² + bx + c

We choose three points (3, -10), (4, -16) and (5, -24) from the table and satisfy the equation to get the values of a, b, and c.

For point (3, -10)

-10 = a(3)² + 3b + c

9a + 3b + c = -10 -------(1)

For point (4, -16)

-16 = a(4)² + 4b + c

16a + 4b + c = -16 ------(2)

For point (5, -24)

-24 = a(5)² + 5b + c

25a + 5b + c = -24 -----(3)

Equation (1) - equation (2)

(9a + 3b + c) - (16a + 4b + c) = -10 + 16

-7a - b = 6

7a + b = -6 ------(4)

Equation (2) - equation (3)

(16a + 4b + c) - (25a + 5b + c) = -16 + 24

-9a - b = 8

9a + b = -8 -------(5)

Equation (4) - Equation (5)

(7a + b) - (9a + b) = -6 + 8

-2a = 2

a = -1

From equation (4),

-7(1) + b = -6

b = -6 + 7

b = 1

From equation (1)

9(-1) + 3(1) + c = -10

-9 + 3 + c = -10

c = -10 + 6

c = -4

Therefore, the value of constant term c is (-4).

5 0
3 years ago
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