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2 years ago

HELP PLEASE!!!!!!!!!!!!

Mathematics

1 answer:

Alex Ar [27]2 years ago

4 0

I dont see a graph??

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17. Fruit drink A consists of 6% pure fruit juice and drink B consists of 15% pure fruit

grigory [225]

Answer: drink A should be mix in all while B should be 3% only

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Need more help again

Fittoniya [83]

Answer:

QSR = 68º

Step-by-step explanation:

TSQ + QSR = TSR

15x + (10x - 2) = 173

Combine like terms

25x = 175

Divide both sides by 25

x = 7

QSR = 10x - 2

QSR = 10(7) - 2

QSR = 68º

5 0

3 years ago

A(x) = -0.015x^3+1.05xA ( x ) = − 0.015 x 3 + 1.05 x gives the alcohol level in an average person's blood x hrs after drinking 8

goldfiish [28.3K]

Answer:

Yes

Step-by-step explanation:

the function that gives the alcohol level is:

$A ( x ) = - 0.015 x^3 + 1.05 x$

where x is the number of hours.

we need to know if after 4 hours an average person is legally drunk, thus:

$x=4$

and we substitute this in the function:

$A ( 4 ) = - 0.015 (4)^ 3 + 1.05(4)$

solving these operations we obtain:

$A(4)=-0.015(64)+4.2\\A(4)=-0.96+4.2$

$A(4)=3.24$

the alcohol level after 4 hours is 3.24.

Since a person is considered to be legally drunk if the level exceeds 1.5, and we obtained 3.24 which is greater than 1.5, a person who has been drinking for 4 hours under the conditions indicated by the problem would be considered legally drunk.

5 0

2 years ago

NEED HELP ASAP!!

Sindrei [870]

The answer is B (the quotient of four time some number and six)

7 0

2 years ago

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The half-life of cesium-137 is 30 years. Suppose we have a

VMariaS [17]

(a) If $y(t)$ is the mass (in mg) remaining after $t$ years, then $y(t) = y(0) e^{kt}=100e^{kt}.$

$y(30) = 100 e^{30k} = \frac{1}{2}(100) \implies e^{30k} = \frac{1}{2} \implies k = -(\ln 2) /30 \implies \\ \\
y(t) = 100e^{-(\ln 2)t/30} = 100 \cdot 2^{-t/30}$

(b) $y(100) = 100 \cdot 2^{-100/30} \approx \text{9.92 mg}$

(c)
$100 e^{- (\ln 2)t/30} = 1\ \implies\ -(\ln 2) t / 30 = \ln \frac{1}{100}\ \implies\ \\ \\
t = -30 \frac{\ln 0.01}{\ln 2} \approx \text{199.3 years}$

3 0

2 years ago