All categories

3 years ago

8x-14=26 what is the anwser

Mathematics

2 answers:

Eva8 [605]3 years ago

4 0

X=1.5 is the correct answer to this equation

KiRa [710]3 years ago

4 0

8x=26+14
8×=40
×=40÷8
×=5

You might be interested in

What is the average of 4 1/2, 3 1/3, 2, 2 1/6? I will give brainleist.

kotykmax [81]

Answer:

Step-by-step explanation:

4 1/2 + 3 1/3 + 2 + 2 1/6

= 4 3/6 + 3 2/6 + 2 + 2 1/6

= 4 + 3 + 2 + 2 + 3/6 + 2/6 + 1/6

= 11 + 6/6

= 11 + 1

= 12

There are 4 numbers, therefore average = 12 ÷ 4 = 3

7 0

2 years ago

How many solutions would it have none infinite or one ?

Jlenok [28]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

This is due Monday but I want to get it done today, because the teacher is still going to check it, and I have practicly nothing

Crank

You don’t have anything attached:(

7 0

3 years ago

Iliana was part of a group that was working on changing 0.4 repeated to a fraction. Each member of the group had a different ans

amm1812

$\bf 0.444444444\overline{4}\impliedby \textit{and keeps on going}\\\\
-------------------------------\\\\
\textit{let's say }\boxed{x=0.444444444\overline{4}}\quad \textit{ thus }10\cdot x=4.44444444\overline{4}
\\\\\\
\textit{wait a minute! }4.44444444\overline{4}\textit{ is really just }4+0.444444444\overline{4}$

$\bf \textit{but we know }x=0.444444444\overline{4} \textit{ so then }4+0.444444444\overline{4}=\boxed{4+x}
\\\\\\
\textit{wait a second! }10\cdot x\implies 10x=4.444444444\overline{4}=4+x
\\\\\\
thus\qquad 10x=4+x\implies 10x-x=4\implies 9x=4\implies \boxed{x=\cfrac{4}{9}}$

you can check in your calculator.

anyhow, to get the "recurring decimal to fraction", you start by setting to some variable, "x" in this case, then move the repeating part to the left of the point by multiplying it by some power of 10, and then do the equating.

3 0

3 years ago

Read 2 more answers

write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.

5 0

3 years ago