3 years ago

Find the distance by taking the absolute value of the difference of the two numbers -12 and 19

Mathematics

1 answer:

Oksi-84 [34.3K]3 years ago

5 0

You'd add 19 and 12 giving you 31

You might be interested in

3(4-2) help please thank you

Talja [164]

Answer: 6

Explanation:
3(4-2) = 3(2) = 6

6 0

3 years ago

Read 2 more answers

A triangle has side lengths measuring 3x cm, 7x cm, and hcm Which expression describes the possible values of h, in cm?

GREYUIT [131]

Answer:

OLX is your answer if 10 x

4 0

3 years ago

A rectangle has an area of 19.38 cm2. When both the length and width of the rectangle are increased by 1.50 cm, the area of the

Lostsunrise [7]

Answer: $5.7\ cm$

Step-by-step explanation:

Given

Rectangle has an area of $19.38\ cm^2$

Suppose rectangle length and width are $l$ and $w$

If each side is increased by $1.50\ cm$

Area becomes $A_2=35.28\ cm^2$

We can write

$\Rightarrow lw=19.38\quad \ldots(i)\\\\\Rightarrow (l+1.5)(w+1.5)=35.28\\\Rightarrow lw+1.5(l+w)+1.5^2=35.28\\\text{use (i) for}\ lw\\\Rightarrow 19.38+1.5(l+w)=35.28-2.25\\\Rightarrow l+w=9.1\quad \ldots(ii)$

Substitute the value of width from (ii) in equation (i)

$\Rightarrow l(9.1-l)=19.38\\\Rightarrow l^2-9.1l+19.38=0\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{(-9.1)^2-4(1)(19.38)}}{2\times 1}\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{5.29}}{2}\\\\\Rightarrow l=\dfrac{9.1\pm2.3}{2}\\\\\Rightarrow l=3.4,\ 5.7$