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PilotLPTM [1.2K]

3 years ago

9

Tell whether the function represents exponential growth or exponential decay. Identify the percent increase or decrease. Then gr

aph the function. y=5^x

1 answer:

inysia [295]3 years ago

4 0

The percent is 4% hope it is

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All of the following are equivalent except _____.<br> 4 · y<br> 4 + y<br> 4y<br> (4)(y)

DENIUS [597]

The answer is: [B]: 4 + y .
______________________
The rest of the answer choices given are all different ways to express, or are equivalents of, the number "4", multiplied by the variable, "y".
______________________

5 0

3 years ago

Read 2 more answers

How to convert improper fractions to a mixed numeral

Deffense [45]

Say for example, you have the improper fraction 18/7
The first step is to find out how many times 7 fits into 18
This looks like flipping the fraction, 7/18 = 2 whole times with 4 parts left over
The 4 left over gets put in front of the original number (7) to give 2 4/7

To check if you have done it correctly, multiply the whole number (2) by the denominator ( 7 ) and then add the numerator.
2 x 7 = 14 + 4 = 18

Hope this helps

4 0

3 years ago

Read 2 more answers

What is the y intercept of 3x62-6x+21=0

TiliK225 [7]

Answer:

(0, 207)

Step-by-step explanation:

We calculate y-intercept by substituting 0 for x.

The equation becomes:

3*62-6*(0)+21=207

y intercept is (0, 207)

3 0

3 years ago

Write the equation of a line that goes through the points (-7,10) and (1,4)

Bad White [126]

I’m pretty sure the answer is y=3/4x-13/4

8 0

3 years ago

Read 2 more answers

If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

4 0

3 years ago