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3 years ago

PLEASE HELP!!! How many solutions does the system have?

Mathematics

1 answer:

Sholpan [36]3 years ago

4 0

Answer:

Infinitely many.

Step-by-step explanation:

3x + 2y = 4

-6x - 4y = -8

Note that the second equation is obtained from the first by multiplying the first one by -2. Thus the graph of the 2 equations is the same line and there are infinitely many solutions.

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3 years ago

Need help with equations with variables on both sides, giving 100 points, tysm if you help :) 6 (2x-4) = 8(x +2) A. 3 B. 10 C. 2

scoundrel [369]

Answer:

B. 10

C. All real numbers.

Step-by-step explanation:

6 (2x-4) = 8(x + 2)

Distribute the numbers outside of the factors:

12x - 24 = 8x + 16

Subtract both sides by '8x'

12x - 8x - 24 = 8x - 8x + 16

4x - 24 = 16

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3(4p - 2) = -6(1 - 2p)

Distribute the numbers similarly to the example before:

12p - 6 = -6 + 12p

Subtract '12p' from both sides:

12p - 12p -6 = -6 + 12p - 12p

0 -6 = -6

Add '6' to both sides:

0 - 6 + 6 = -6 + 6

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Therefore, the solution consists of all real numbers.

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Read 2 more answers

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3>Let's understand the concept:-</h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3>Required Answer:-</h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

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3 years ago