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3 years ago

What does a diagonal line look like????

Mathematics

2 answers:

zepelin [54]3 years ago

5 0

A diagonal line looks like this \ or /

* an x is made up of two diagonal lines.

lisabon 2012 [21]3 years ago

3 0

In a simple word, a line that is crooked, or slanted. Like this: /

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If the temperature on the ground is 70 F then

harina [27]

Answer: try to (x) each one by the answer to -0.0035 +70

Step-by-step explanation:

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4 years ago

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five percent of a batch of fuel was found faullty if 27 fuses were faulty how many were inspected the price of internet service

LenKa [72]

There are 540 fuses in total and the original price of the internet is $55.91

<h3>How to work with percentages?</h3>

a) We know that 5% of the fuses are faulty, and there are 27 faulty fusses, then we can write:

5% = 27

100% = x

We want to find x, to do that, we can write:

x = 27*(100%/5%) = 27*20 = 540

There are 540 fuses in total.

b) Now we know that the original price is P, and when we decrease it by 7%, then new price is $52.00

Then we can write:

$52.00 = P*(1 - 7%/100%)

$52.00 = P*(0.93)

$52.00/0.93 = P = $55.91

The original price is $55.91

If you want to learn more about percentages:

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2 years ago

Which symbol correctly compares the fractions below?<br> 2/5 ? 7/10

aliya0001 [1]

Answer:

2/5 < 7/10

I think this is it

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3 years ago

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8) slope and y-intercept from graphs

Artyom0805 [142]

Answer:slope-0.75 y intercept 1

Step-by-step explanation:

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3 years ago

Solve the rational equation x divided by 6 equals x squared divided by quantity x minus 1 end quantity, and check for extraneous

pychu [463]

Answer:

$x = 0$ and $x = \textrm{negative one fifth}$

Step-by-step explanation:

Given:

The rational equation to solve is given as:

$\frac{x}{6}=\frac{x^2}{x-1}
$

We use the cross multiplication rule and multiply the numerators one side with the denominators of the other side. Therefore,

$x\times (x-1)=6\times x^2\\x^2-x=6x^2\\\textrm{Bringing all variables to the right side, we get:}\\6x^2-x^2+x=0\\5x^2+x=0\\x(5x+1)=0\\x=0\ or\ 5x+1=0\\x=0\ or\ x=-\frac{1}{5}
$

Now, for $x=0$ , the rational equation is equal to 0. So, $x=0$ is a solution.

Also, $x=-\frac{1}{5}$ is also a solution as the equation's denominator is defined at that point.

So, no extraneous solution.

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4 years ago