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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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The area of a triangular block is 49 square inches. If the base of the triangle is twice the height, how long are the base and t
andrezito [222]
1) Formula: area = height * base / 2

2) Call x the height:

height = x

base = 2x

3) State the equation:

(x)(2x) = 49 in^2

4) Solve the equation:

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x^2 = (49/2) in^2

x = √(49/2) in

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5) Solution:

height = 3.5 √2 in and base = 7√2 in

6) Verification: area = (7√2 in) (3.5√2 in) = 49 in^2

Answer: height = 3.5√2 in and base = 7√2 in.
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3 years ago
What is the side length a in the triangle below?
grigory [225]

Answer:

a=5

Step-by-step explanation:

a^2+b^2=c^2

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13^2-12^2=a^2

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8 0
2 years ago
Need help pls answer asap. Sorry the photo quality is bad.
castortr0y [4]

Answer:

Step-by-step explanation:

1 + 3 = 4

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x = -1 + 2 = 1

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Δy =  ¼(8 - 2) = 1½

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y = 2 + 1½ = 3½

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3 years ago
What is 6.3-2(1.5c+4.1)
ASHA 777 [7]
Expand\;-2\left(1.5c+4.1\right)
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Simplify\;-2\cdot \:1.5c-2\cdot \:4.1 \ \textgreater \  \mathrm{Multiply\:the\:numbers:}\:2\cdot \:1.5=3
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\mathrm{Multiply\:the\:numbers:}\:2\cdot \:4.1=8.2 \ \textgreater \  -3c-8.2 \ \textgreater \  6.3-3c-8.2

\mathrm{Subtract\:the\:numbers:}\:6.3-8.2=-1.9 \ \textgreater \  -3c-1.9

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Substitute -1 for the variable x.

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3 years ago
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