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3 years ago

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What is the fraction if you have $.95 And for the coins are twice as much as the rest how can I construct a Mac argument could j

ustify and for the coins are twice as much as the rest how can I construct a math argument to justify the conjecture of 11 nickels and 4 dimes

1 answer:

DaniilM [7]3 years ago

6 0

I really need five points

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Loss of traction to the rear tires can result in __________ .

Andru [333]

I believe it would be B. Only because i know that brakes are on the back tires but I'm only 14. so I don't know if different cars would have breaks in the back.

HOPE THIS HELPS!!!

6 0

3 years ago

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Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]

Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are $\frac{14}{3}$ and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

$f(x) = \frac{3x-2}{x+1}$
$g(x)=x+5$
f[ -4 ] = ?
g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

$f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}$

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

$g(\frac{3x-2}{x+1}) = (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) = \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) = 13$

Therefore, the value of f[ -4 ] and g°f[-2] are $\frac{14}{3}$ and 13 respectively.

Learn more about composite functions here: brainly.com/question/20379727

#SPJ1

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2 years ago

Solve for x:<br> 4x + 10 = 34

VladimirAG [237]

Answer:

x=6

Step-by-step explanation:

4x+10=34

4x=24

x=6

7 0

3 years ago

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Please I need help badly and quickly, pls do not answer if you just want the points. I need an accurate answer for this. Cheers

natita [175]

Answer

Q11 is 50,3

Q12, 23,19

Step-by-step explanation:

7 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

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