PLEASE HELP ME ON THIS AND SHOW UR WORK THX

Mathematics

1 answer:

swat323 years ago

4 0

When an exponential function is given as
f(x)=ab^x, and
knowing that b^0=1, that means that
f(0)=ab^0=a value of function when x=0, or the initial value.
In the case of
f(x)=20.b^x, f(0)=20.b^0 = 20 (initial value)

If b=1.2, then the hourly growth rate is (1.2-1)=0.2=20%.
If b=1.85, then the hourly growth rate is (1.85-1)=0.85=85%.

You might be interested in

Use the distributive property to simplify the expressions. 1B. -3(9x-q) 2B. 2(12+5p) 3b. -7(-8y+7) 4b. 10(2+7c) 5b. -8(7+11k) 6b

Debora [2.8K]

1B: -3(9x-q)
-27x + 3q

Step 1: Distribute the -3 to the 9x to get -27x.
Step 2: Distribute the -3 to the -q.

2B: 2(12+5p)
24 + 10p

Step 1: Distribute the 2 to the 12 to get 24.
Step 2: Distribute the 2 to the 5p to get 10p.

3B: -7(-8y+7)
56y - 49

Step 1: Distribute the -7 to the -8y to get 56y.
Step 2: Distribute the -7 to the 7 to get -49.

4B: 10(2+7c)
20 + 70c

Step 1: Distribute the 10 to the 2 to get 20.
Step 2: Distribute the 10 to the 7c to get 70c.

5B: -8(7+11k)
-56 - 88k

Step 1: Distribute the -8 to the 7 to get -56.
Step 2: Distribute the -8 to the 11k to get -88k.

6B: -(3-7u)
-1(3-7u)
-3 + 7u

Step 1: Place a 1 after the negative symbol to symbolize -1.
Step 2: Distribute the -1 to the three to get -3.
Step 3: Distribute the -1 to the -7u to get 7u.

7B: -6(5p + s)
-30p - 6s

Step 1: Distribute the -6 to the 5p to get -30p.
Step 2: Distribute the -6 to the s to get -6s.

:) :D

8 0

3 years ago

Could someone help me, please?

k0ka [10]

We can rewrite the expression under the radical as

$\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4$

then taking the fourth root, we get

$\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|$

Why the absolute value? It's for the same reason that

$\sqrt{x^2}=|x|$

since both $(-x)^2$ and $x^2$ return the same number $x^2$ , and $|x|$ captures both possibilities. From here, we have

The absolute values disappear on all but the $b$ term because all of $\dfrac32$ , $a^2$ and $c^4$ are positive, while $b^3$ could potentially be negative. So we end up with