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frosja888 [35]
3 years ago
15

A candymaker makes 500 pounds of candy per week, while his large family eats the candy at a rate equal to Q(t)/10 pounds per wee

k, where Q(t) is the amount of candy present at time t.A) Find Q(t) for t > 0 if the candymaker has 250 pounds of candy at t=0.B) Find lim Q(t) t-> infinity.
Mathematics
1 answer:
elena55 [62]3 years ago
5 0

Answer:

Step-by-step explanation:

250 pounds

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Is 3.14596387 a rational number?
zaharov [31]
No it is an irrational number
8 0
3 years ago
Read 2 more answers
42 /48 in simplest form
zloy xaker [14]

<span>First we need to find the GCF (greatest common factor)</span>

<span>Factors of 42 are:
1, 2, 3, 6, 7, 14, 21, 42</span>

<span>Factors of 48 are:
1, 2, 3, 4, 6, 8, 12, 16, 24, 48</span>

<span>GCF is 6</span>

<span>To simplify you divide numerator and denominator into the GCF</span>

<span>42 ÷ 6 = 7</span>

<span>48 ÷ 6 = 8</span>

<span>42/48 = 7/8</span>

<span>Hope this helps. :)
</span>

6 0
3 years ago
What is the product of 7/8×1/2
Gelneren [198K]

\frac{7}{8}   \times  \frac{1}{2}

=  \frac{7 \times 4}{8}

=  \frac{28}{8}

=  \frac{ \cancel{28}}{ \cancel{8}}

= \frac{14}{4}

=  \frac{ \cancel{14}}{ \cancel{4}}  =  \frac{7}{2}

<h2>The required answer is 7/2 </h2>

3 0
3 years ago
Cual es el complementario de un angulo
jonny [76]

Answer:

La suma de los ángulos complementarios es 90 °

Step-by-step explanation:

3 0
3 years ago
How do you solve this equation?
Travka [436]
Exponentail thingies

easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0

now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero

5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1


5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4

x=log₅1 and/or log₅4







second quesiton

same thing

1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero

2^x-8=9
2^x=8
x=3

2^x-2=0
2^x=2
x=1

x=3 or 1







first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3


7 0
3 years ago
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