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sergeinik [125]
3 years ago
14

Which is true regarding the system of equations? x minus 4 y = 1. 5 x minus 20 y = 4.

Mathematics
2 answers:
enot [183]3 years ago
7 0

Step-by-step explanation:

x - 4y = 1.5 \:  \:  \:  \: ...(1) \\ x - 20y = 4 \:  \:  \:  \: ...(2)

from equation (2) we get,

x = 20y + 4

now put the value of x = 20y + 4 in equation (1)

20y + 4 - 4y = 1.5 \\ 16y = 1.5 - 4 \\ 16y =  - 2.5 \\ y =  -  \frac{25}{16 \times 10}   \\ \\ y =  -  \frac{5}{32}

now put the value of y in equation (2) we get

x - 20( -  \frac{5}{32} ) = 4 \\  \\ x +  \frac{100}{32}  = 4 \\  \\ x +  \frac{25}{8}  = 4 \\  \\ x = 4 -  \frac{25}{8}  \\  \\ x =  \frac{32 - 25}{8}  \\  \\ x =  \frac{7}{8}

goldfiish [28.3K]3 years ago
3 0

Answer:

No solution

Step-by-step explanation:

x - 4y = 1 --> (1)

5x - 20y = 4 --> (2)

y = (¼)x - ¼ --> (1)

y = (¼)x - ⅕ --> (2)

Since these are 2 parallel lines with different y-intercepts, they will never meet

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Answer: No, the answer is 29, this is false.

Step-by-step explanation: 63-x=34

subtract 63 from both sides

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EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

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Answer,

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So, the total time taken to reach the destination is 2:45+4:00=6:45hour

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11 POINTS PLZ HELP EMERGENCY
Bad White [126]

Part A

The given expression is:

(-5(1+2i)+3i(3-4i)

We expand to get:

-5-10i+9i-12i^2

Note that i^2=-1

12-5-10i+9i

-5-10i+9i+12

7-i

Part Bi)

The given expression is;

\sqrt{-50}

We simplify to get:

\sqrt{25\times 2\times -1}

\sqrt{25\times} \sqr{2}\times \sqrt{-1}

Note that:

\sqrt{-1}=i

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This is now of the form a+bi, where a=0,b=5\sqr{2}.

This explains why it is a complex number;

Part bii)

The given expression is :

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We simplify to get

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\sqrt{-25}=\sqrt{25}\times \sqrt{-1}

\sqrt{-25}=5\times i

\sqrt{-25}=5i

Part C

We want to simplify:

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We rewrite in terms of i^2

i^{22}=(i^2)^{11}

i^{22}=(-1)^{11}

i^{22}=-11

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