Answer:
![The\:number\:is\:-35](https://tex.z-dn.net/?f=The%5C%3Anumber%5C%3Ais%5C%3A-35)
Step-by-step explanation:
![Let\:x\:be\:the\:number,\:we\:have\:an\:equation:\\\frac{3x}{5}=-21\Leftrightarrow 3x=-105\Rightarrow x=-35](https://tex.z-dn.net/?f=Let%5C%3Ax%5C%3Abe%5C%3Athe%5C%3Anumber%2C%5C%3Awe%5C%3Ahave%5C%3Aan%5C%3Aequation%3A%5C%5C%5Cfrac%7B3x%7D%7B5%7D%3D-21%5CLeftrightarrow%203x%3D-105%5CRightarrow%20x%3D-35)
Answer:
10 degrees
Step-by-step explanation:
Scale factor is multiplying, so to undo you divide
20/2 = 10
m<R = 10
Answer:
The volume of a bar of gold with a mass of 66.5 grams is ![3.5cm^{3}](https://tex.z-dn.net/?f=3.5cm%5E%7B3%7D)
Step-by-step explanation:
Given , mass(in grams) ∝ Volume (in cm3)
Here, volume of 95 grams bar = 5cm3
So, thevolume of 1 gram bar =
= ![\frac{1}{19} cm^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B19%7D%20%20cm%5E%7B3%7D)
⇒The volume of a bar with weight 66.5 grams = ![\frac{1}{19} \times 66.5 = 3.5 cm^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B19%7D%20%20%5Ctimes%2066.5%20%3D%20%203.5%20cm%5E%7B3%7D)
So, the volume of a bar of gold with a mass of 66.5 grams is ![3.5cm^{3}](https://tex.z-dn.net/?f=3.5cm%5E%7B3%7D)