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alekssr [168]
3 years ago
8

Help please !!!!!! Find Y

Mathematics
1 answer:
Darya [45]3 years ago
7 0

Answer:

Turn it around

Step-by-step explanation:

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Shaggy earned an average of $112 per day for solving mysteries last week Monday through Friday. What was Shaggy total earnings l
kati45 [8]

112 x 5=560

Hay que multiplicar 112 x 5 y listo.

4 0
3 years ago
Jillian ran 40 miles in one week. Her friend Lacey ran the distance that Jillian ran. How many miles did Lacey run? Express your
vichka [17]

I'm SURE you left something out of the space between "ran" ... and ...
"the distance". But here in the Math category, it's our job to answer the
questions, not to fix them.

The question tells us that Jillian ran 40 miles, and that Lacey ran
the distance that Jillian ran.  We can logically conclude, therefore,
that Lacey ran 40 miles.

3 0
4 years ago
14. What is the area of a rectangle if the dimensions are:<br> 5y2 in and 8y3 in
ololo11 [35]

Answer:

40 y^4 unit sq.

Step-by-step explanation:

Area of a rectangle = length x breadth

                                =  5y^2 x 8y^3

                                = 40 y^5 unit sq.    [Exponents get added when multiplied]

4 0
3 years ago
BRAINLIESTTTT!!!!
Tresset [83]
We have that
x²<span> − 2x − 3 = 0

Step 1
</span>isolate the variable terms 
<span>x² - 2x = 3 

</span>Step 2
complete the square 
<span>x² - 2x + 1² = 3 + 1² </span>
<span>(x-1)² = 4 
</span>
the answer is the option (x-1)² = 4 
8 0
4 years ago
A population of 490 bacteria is introduced into a culture and grows in number according to the equation below, where t is measur
Lera25 [3.4K]

Answer:

Rate of growth of bacteria when t=2 is 3.09 bacteria/hour

Step-by-step explanation:

As equation is not given so considering the Equation of growth of bacteria as

P=490(1+\frac{4t}{50+t^{2}})

We have to find the rate at which population is growing. To do so differentiate above equation w.r.to 't'

\frac{dP}{dt}=\frac{d}{dt}490(1+\frac{4t}{50+t^{2}})\\\\\frac{dP}{dt}=490(\frac{d}{dt}(1)+\frac{d}{dt}(\frac{4t}{50+t^{2}}))\\\\\frac{dP}{dt}=490(0+\frac{4(50+t^{2})-(4t)(2t)}{(50+t^{2})^{2}})\\\\\frac{dP}{dt}=490(\frac{200+4t^{2}-8t^{2}}{(50+t^{2})^{2}})\\\\\frac{dP}{dt}=490(\frac{4(50-t^{2})}{(50+t^{2})^{2}})\\\\at\,\,t=2hours\\\\\frac{dP}{dt}=490(\frac{4(50-(2)^{2})}{(50+(2)^{2})^{2}})\\\\\\\frac{dP}{dt}=490(\frac{4(50-4)}{(50+4)^{2}})\\\\=3.09

Rate of growth of bacteria when t=2 is 3.09 bacteria/hour

8 0
3 years ago
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