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3 years ago

Which statement is true about the equations-3x 4y 12 and tx 1?

Mathematics

1 answer:

nydimaria [60]3 years ago

7 0

We need to solve -3x+4y=12 for x

Let's start by adding -4y to both sides

-3x+4y-4y=12-4y

-3x=-4y+12

x = (-4y+12)/-3

x= 4/3 y -4

Now substitute 4/3 y -4 for x in 1/4 x - 1/3 y =1

1/4 x -1/3 y =1

1/4 (4/3 y -4) -1/3 y =1

Use the distributive property

(1/4)(4/3 y) + (1/4)(-4) -1/3 y =1

1/3 y -1 - 1/3 y =1

Now combine like terms

(1/3y -1/3y) + (-1) =1

= -1

-1 = 1

Now add 1 to both sides

0=2

So there are no Solutions

The answer is C

I hope that's help Will :)

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3 years ago

Solve the x <br> 1/3x + 5> 10

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Answer:

The solution to the inequality is:

$\frac{1}{3}x+5>10\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>15\:\\ \:\mathrm{Interval\:Notation:}&\:\left(15,\:\infty \:\right)\end{bmatrix}$

The solution graph is also attached below.

Step-by-step explanation:

Given

We are given the expression

$\frac{1}{3}x+5>10$

To determine

Solve for x

Given the expression

$\frac{1}{3}x+5>10$

Subtract 5 from both sides

$\frac{1}{3}x+5-5>10-5$

Simplify

$\frac{1}{3}x>5$

Multiply both sides by 3

$3\cdot \frac{1}{3}x>5\cdot \:3$

Simplify

$x>15$

Thus, we conclude that:

$x>15$

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$\frac{1}{3}x+5>10\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>15\:\\ \:\mathrm{Interval\:Notation:}&\:\left(15,\:\infty \:\right)\end{bmatrix}$

The solution graph is also attached below.

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A triangle, ABC, has coordinates A (-2, 0), B (-2, 5) and C (-4, 0). What is the length of the shortest side?

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Answer: 2 units.

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a^2 + b^2 = c^2, where a and b are the legs (the other sides of the triangle) and c is the hypotenuse.

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√ 29 ≈ 5.38, and that is the length of the hypotenuse.

The shortest side is the bottom side, which is 2 units. Therefore, that is your answer.

Let me know if this is right! :)

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