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attashe74 [19]
3 years ago
14

4/8 times 3/4 simplified pls and thank you

Mathematics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

3/8

Step-by-step explanation:

divide out the 4's and you are left with 3/8

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Step-by-step explanation:

\cos \: 60 \degree =  \frac{x}{11}  \\  \\  \therefore \:  \frac{1}{2}  =  \frac{x}{11}  \\  \\   \therefore \: x =  \frac{11}{2}  \\  \\   \:  \:  \:  \:  \:  \: \huge \red{ \boxed{\therefore \: x =  5.5}} \\

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3 years ago
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These two rectangles are similar. Find the missing dimension. (I need an answer ASAP, I also need a step by step answer)
nevsk [136]
20/5=4 so 5x4 which means 2x4=8 so x is 8 hopes this helps
7 0
3 years ago
Please help, I didn't learn this in class!
NemiM [27]
Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)


Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)


Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1, y = 1, z = 0

7 0
3 years ago
How do solve this equation step by step
Kryger [21]

Answer: 15n³-105n²+2n+16/6n²-42n

Step-by-step explanation:

n+8/3n²-21n +5n/2

2(n+8)+5n(3n²-21n)/2(3n²-21n)

2n+16+15n³-105n²/6n²-42n

15n³-105n²+2n+16/6n²-42n

7 0
3 years ago
Using the graph above, create a system of inequalities that only contains points B and C in the overlapping shaded regions. Expl
k0ka [10]

1. The system of inequality is

y < 1.5x - 3

y < -2x/3 + 4



2. A is (2,-3)

E is (3,1)

-3 < 1.5(2) - 4

-3 < -1

1 < -2(3)/3 + 4

1 < 2



3. You can graph the inequality and see which schools inside the zone. Schools C and B are the schools he is allowed to attend.



hope it helps

8 0
3 years ago
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