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Lunna [17]
3 years ago
9

Negative Two thirds (3x-4)+3x= five sixths

Mathematics
2 answers:
slamgirl [31]3 years ago
5 0

Step-by-step explanation:

Very easy, replace x with everything

equation 1: -2/3(3*-19/6)-4]+3*-19/6=5/6

Equation 2: -2/3 (3*-11/6)-4]+3*-11/6=5/6

equation 3: -2/3 (3*21/6)-4]+3*21/6=5/6

Equation 4: -2/3 (3*29/6)-4]+3*29/6=5/6

Solve all 4 equations

Cheers! Mark brainliest if helpful enough.

Olenka [21]3 years ago
4 0

Answer:

x =  -  \frac{11}{6}

Step-by-step explanation:

-  \frac{2}{3} (3x - 4) + 3x =  \frac{5}{6}

- 2x +  \frac{8}{3}  + 3x =  \frac{5}{6}

( - 2x + 3x) +  \frac{8}{3}  =  \frac{5}{6}

x +  \frac{8}{3 }  =  \frac{5}{6}

x =  \frac{5}{6}  -  \frac{8}{3}

x =  \frac{5}{6}  -  \frac{16}{6}

x =  \frac{5 - 16}{6}

x =   - \frac{11}{6}

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What is the perimeter of the triangle shown on the coordinate plane,to the nearest tenth of a unit ?
ollegr [7]

Answer:

25.6 units

Step-by-step explanation:

From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}

where

(x_1,y_1) are the coordinates of the first point

(x_2,y_2) are the coordinates of the second point

- For AB:

d=\sqrt{[1-(-5)]^{2}+(4-4)^2}

d=\sqrt{(1+5)^{2}+(0)^2}

d=\sqrt{(6)^{2}}

d=6

- For BC:

d=\sqrt{(3-1)^{2} +(-4-4)^{2}}

d=\sqrt{(2)^{2} +(-8)^{2}}

d=\sqrt{4+64}

d=\sqrt{68}

d=8.24

- For AC:

d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}

d=\sqrt{(3+5)^{2} +(-8)^{2}}

d=\sqrt{(8)^{2} +64}

d=\sqrt{64+64}

d=\sqrt{128}

d=11.31

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

p=AB+BC+AC

p=6+8.24+11.31

p=25.55

p=25.6

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

6 0
4 years ago
-4+x= -11 please answer it
Slav-nsk [51]

Answer:

x=-7

Step-by-step explanation:

add 4 to both sides. leaving you with x=-7

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