What are the domain and range of fx) = (1/5)^x

Mathematics

1 answer:

a_sh-v [17]2 years ago

6 0

Answer: the domain is all real numbers. The Range is all real numbers greater than zero.

Step-by-step explanation:

this si the same as the graph y=5^-x. Which decreases from left to right and approaches but never touches the x-axis.

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25 POINTS AND BRAINLIEST PLEASE HELP ASAP

dexar [7]

M is a midpoint of BC so:

$M=\left(\dfrac{\boxed{2}\boxed{a}+a}{\boxed{2}},\dfrac{\boxed{0}+b}{2}\right)=\left(\dfrac{\boxed{3}\boxed{a}}{\boxed{2}},\dfrac{\boxed{b}}{\boxed{2}}\right)$

Length of MA:

$MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=
\sqrt{\left(\dfrac{\boxed{3}a}{\boxed{2}}\right)^2+\left(\dfrac{b}{2}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

Length of NB:

$NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\
\sqrt{\left(\dfrac{-3a}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

5 0

3 years ago

Read 2 more answers

Mr mehra spends 30% of his income in food 10% on charity 25% and is left with 15750 what is his salary? find his savings

nalin [4]

Answer:

Mr mehra's salary is 26250

Step-by-step explanation:

suppose,

Mr mehra's salary is x

so,

He spends in food = 30% of x

=(30/100) . x [x%=x/100]

= 30x/100

=3x/10

He spends in charity =10% of x

=10x/100

=x/10

Money left = 15750

so,

x = 3x/10 + x/10 + 15750

⇒x - 3x/10 - x/10 = 15750

⇒(10x-3x-x) /10 = 15750

⇒6x/10 = 15750

⇒6x=157500

⇒x = 157500/6

⇒x = 26250

∴Mr mehra's salary is 26250

5 0

3 years ago

Need help only good answer plis!

Nesterboy [21]

A. The angles at the intersection of the two lines can be proven to be congruent and complementary . so they meet at a right angle and the lines are perpendicular.

<u>Step-by-step explanation:</u>

In above question, In order to find whether AB ⊥ CD, Using compass construction & rounder , keep the tip at A and cut arcs at line CD . Follow the same process again with tip at B and cut arcs at line CD . Do this both sides of Line CD i.e. on left side of AB & on right side of AB. Now, join the intersection points of both side arcs which are intersecting each other. Now, to prove both are right angle to each other i.e. AB ⊥ CD , can be done by proving congruent and complementary , so they meet at a right angle and Hence , the lines are perpendicular i.e. AB is inclined to CD at angle of 90°.