47. You subtract the lowest number from the highest to get your range.
Answer:
<h2>For c = 5 → two solutions</h2><h2>For c = -10 → no solutions</h2>
Step-by-step explanation:
We know
![|a|\geq0](https://tex.z-dn.net/?f=%7Ca%7C%5Cgeq0)
for any real value of <em>a</em>.
|a| = b > 0 - <em>two solutions: </em>a = b or a = -b
|a| = 0 - <em>one solution: a = 0</em>
|a| = b < 0 - <em>no solution</em>
<em />
|x + 6| - 4 = c
for c = 5:
|x + 6| - 4 = 5 <em>add 4 to both sides</em>
|x + 6| = 9 > 0 <em>TWO SOLUTIONS</em>
for c = -10
|x + 6| - 4 = -10 <em>add 4 to both sides</em>
|x + 6| = -6 < 0 <em>NO SOLUTIONS</em>
<em></em>
Calculate the solutions for c = 5:
|x + 6| = 9 ⇔ x + 6 = 9 or x + 6 = -9 <em>subtract 6 from both sides</em>
x = 3 or x = -15
Answer:
do u need other coordinates?
if so B'(1,-1)
C'(1,4)
D'(5,6)
the formula is (-x,y)
hope it helps :)
![\text{Let the product of two natural numbers p and q is 590, and their HCF is 59}\\ \\ \text{we know that the product of LCM and HCF of any two numbers is equal}\\ \text{to the product of the numbers. that is}\\ \\ \text{HCF}\times \text{ LCM}=p\times q\\ \\ \Rightarrow 59 \times \text{LCM}=590\\ \\ \Rightarrow \text{LCM}=\frac{590}{59}\\ \\ \Rightarrow \text{LCM}=10\\ \\ \text{for any two natural numbers, their Least Common Multiple (LCM) is always}](https://tex.z-dn.net/?f=%20%5Ctext%7BLet%20the%20product%20of%20two%20natural%20numbers%20p%20and%20q%20is%20590%2C%20and%20their%20HCF%20is%2059%7D%5C%5C%0A%5C%5C%0A%5Ctext%7Bwe%20know%20that%20the%20product%20of%20LCM%20and%20HCF%20of%20any%20two%20numbers%20is%20equal%7D%5C%5C%0A%5Ctext%7Bto%20the%20product%20of%20the%20numbers.%20that%20is%7D%5C%5C%0A%5C%5C%0A%5Ctext%7BHCF%7D%5Ctimes%20%5Ctext%7B%20LCM%7D%3Dp%5Ctimes%20q%5C%5C%0A%5C%5C%0A%5CRightarrow%2059%20%5Ctimes%20%5Ctext%7BLCM%7D%3D590%5C%5C%0A%5C%5C%0A%5CRightarrow%20%5Ctext%7BLCM%7D%3D%5Cfrac%7B590%7D%7B59%7D%5C%5C%0A%5C%5C%0A%5CRightarrow%20%5Ctext%7BLCM%7D%3D10%5C%5C%0A%5C%5C%0A%5Ctext%7Bfor%20any%20two%20natural%20numbers%2C%20their%20Least%20Common%20Multiple%20%28LCM%29%20is%20always%7D%20)
![\text{greater than their HCF.}\\ \\ \text{but here we can see that }LCM](https://tex.z-dn.net/?f=%20%5Ctext%7Bgreater%20than%20their%20HCF.%7D%5C%5C%0A%5C%5C%0A%5Ctext%7Bbut%20here%20we%20can%20see%20that%20%7DLCM%20%3CHCF%20)
Hence there is no such natural numbers exist.
Answer:
There are 6 adults. <em>My mistake! I had the wrong equation. 3 + 9a = 57.</em>
Step-by-step explanation:
- <u>There is only 1 child in the group.</u>
- Cost for children is 3$.
- <u>The cost for adults is 9$.</u>
- 9 * 6 = 54
- 3 * 1 = 3
- 3 + 9a = 57
- Simplify:
- 6 Adults.