The length of an insect’s wing, drawn at a scale of 50:1, is 15 cm. What is the actual length of the wing?

Mathematics

2 answers:

MAXImum [283]3 years ago

6 0

Answer:

0.3 cm

Step-by-step explanation:

The scale is 50:1 this means if the wing had a real length of 1 cm, in the drawing it will be of 50 cm.

We want to know the real length of the wing of the drawing has a wing with a length of 15cm.

We can express the information we have as follows:

Drawn length Real Length

50cm 1cm

15cm x

Where x is the real length of the wing when the drawing has a length of 15cm.

We can solve the problem by applying rule of three, we must multiply the cross quantities and divide by the remaining amount.

$x=\frac{15*1}{50}=\frac{15}{50} =0.3$

The actual length of the wing is 0.3cm.

MaRussiya [10]3 years ago

3 0

The answer is 750.
15×50=750.

You might be interested in

Simplify -4(3-1)+2

nikdorinn [45]

To solve this you must use the rules of PEMDAS (Parentheses, Exponent, Multiplication, Division, Addition, Subtraction)

First we have parentheses.

-4(3 - 1) + 2

3 - 1 = 2

-4(2) + 2

We have no exponents so go on to multiplication

-4(2) + 2

-4 * 2 = -8

-8 + 2

No division so go to addition

-8 + 2

When adding a positive number with a negative number you will act as if you are subtracting the two numbers, then take the sign of the largest number (disregarding the signs). In this case the largest number is 8 and its sign is negative. Your answer will have a negative sign.

-6

Hope this helped!

~Just a girl in love with Shawn Mendes

7 0

3 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl

aev [14]

$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$