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3 years ago

Is Y=-3x-2 linear or no?

Mathematics

1 answer:

Tcecarenko [31]3 years ago

3 0

Yes!

This would be in-fact a linear equation!

This equation is set up in y=mx+b form, therefore we know it is a linear equation.

<h3>Example of a linear equation:</h3>

y=12x+2

The 12 is the mx in the equation.

And the 2 is the B in the equation! :)

That is how we solve that.

Any questions?

Have a great day!

You might be interested in

Correct answer get the BRAINLIEST ANSWER!

OLga [1]

Here is how I completed the problem:

I started by creating a ratio:

$\frac{x}{0.18} + \frac{360}{0.10} = \frac{x+360}{0.15}$

<em>Where 'x' is the desired amount of the 18% solution of sulfuric acid.

</em>From here, I solved for 'x':
<em>
</em> $\frac{0.10x}{0.018} + \frac{360(0.18)}{0.018} = \frac{x+360}{0.15}$
$\frac{0.10x}{0.018} + \frac{64.8}{0.018} = \frac{x+360}{0.15}$
$\frac{0.10x+64.8}{0.018} = \frac{x+360}{0.15}$
$0.15(0.10x+64.8) = 0.018(x+360)$
$0.015x+9.72 = 0.018x+6.48$
$0.003x = 3.24$
$x = 1080$

∴You will need to add 1080ml of 18% Sulfuric Acid in order to obtain a 15% solution.

8 0

3 years ago

I’d really appreciate if you’d also give a brief explanation:)

ArbitrLikvidat [17]

Answer:

Only area of PQR can be calculated.

Step-by-step explanation:

In PQR, we have both the base (24 cm) and the height (15cm) which is perpendicular to the base. Therefore using these two measurements, we can take

$\frac{1}{2} \times 24 \times 15$

This is equal to 180 cm².

3 0

3 years ago

Find a cubic function with the given zeros.

Fed [463]

Answer:

The correct option is D) $f(x) = x^3 + 2x^2 - 2x - 4$ .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x + 4$ .

$f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

$f(x) = x^3 + 2x^2 + 2x - 4\\f(x) = (-2)^3 + 2(-2)^2 + 2(-2) - 4\\f(x) =-8+2(4)-4-4\\f(x) =-8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 - 2x^2 - 2x - 4$ .

$f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

8 0

4 years ago

One-third (x minus 2) = one-fifth (x + 4) + 2?

Diano4ka-milaya [45]

Answer:

x = 26

Step-by-step explanation:

6 0

4 years ago

The answer to this problem

horrorfan [7]

Work backwards with pythagoras' theorem:
c^2=a^2+b^2
Here, a and b both equal to x. Therefore, you can write this as 8^2=2x^2. So 64=2x^2. Sqrt on both sides to give you 8=2x, hence x=4. However I must warn you that this working out may have a flaw, as I found that x actually equals to 5.66

7 0

4 years ago