The width of the insect is 0.17 cm and the height of the insect is 0.125 cm.
Step-by-step explanation:
Step 1:
The scale factor is given by dividing the measurement after scaling by the same measurement
The given drawing of the insect is larger than the actual insect.
The scale factor is 20: 1 which means 20 cm on the drawing is 1 cm on the actual insect.
Step 2:
To obtain the actual insect's dimensions, we divide the dimensions of the drawing by 20.
The width of the actual insect ![= \frac{3.4}{20} = 0.17.](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B3.4%7D%7B20%7D%20%3D%200.17.)
The height of the actual insect ![= \frac{2.5}{20} = 0.125.](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B2.5%7D%7B20%7D%20%3D%200.125.)
The width of the insect is 0.17 cm and the height of the insect is 0.125 cm.
We aer goint to eliminate the x's
multiply first by -5 and 2nd by 2
so that is B
Answer: The value of x is 6.
Step-by-step explanation:
TO find the volume of a cube you cube any side length. so if the side length is 6.
6^3 = 216
To find the surface area of a cube you will square one of the side lengths and multiply it by 6.
so the side length is 6 and 6 squared is 36
36*6 = 216
Yes and they will always be coplanar.
Answer:
<em>Riko is wrong because she uses the wrong formula.</em>
Step-by-step explanation:
Given the coordinate points (-2, -3) and (2, -5)
We are to find the distance between the two points
Using the formula:
D = √(x2-x1)²+(y2-y1)²
From the coordinates x1 = -2, y1 = -3, x2 = 2 and y2 = -5
Substitute
D = √(2-(-2))²+(-5-(-3))²
D= √(4)²+(-2)²
D = √16 + 4
D = √20
D = √4*√5
D = 2√5
<em>This shows that Riko is wrong because she uses the wrong formula.</em>