Suzanne can work 10 math problems in 15 minutes. How many can she work in an hour and a half?

Mathematics

1 answer:

satela [25.4K]3 years ago

7 0

Hey there!

10-15
x-90
so 15x=900
x=60

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Expand and simplify 4(2x +7) – 3(x + 5)

Vikentia [17]

Answer:

$4\left(2x+7\right)-3\left(x+5\right)=5x+13$

Step-by-step explanation:

Given the expression

$4\left(2x+7\right)-3\left(x+5\right)$

<u>Expand and simplify the expression</u>

$4\left(2x+7\right)-3\left(x+5\right)$

Expand 4(2x + 7) = 8x + 28

$4\left(2x+7\right)-3\left(x+5\right)=8x+28-3\left(x+5\right)$

Expand -3(x + 5) = -3x - 15

$=8x+28-3x-15$

Group like terms

$=8x-3x+28-15$

Add similar elements: 8x - 3x = 5x

$=5x+28-15$

Simplify

$=5x+13$

Therefore, we conclude that:

$4\left(2x+7\right)-3\left(x+5\right)=5x+13$

6 0

3 years ago

A sample of tritium-3 decayed to 94.5% of its original amount after a year.

Zolol [24]

(a) If $y(t)$ is the mass after $t$ days and $y(0) = A$ then $y(t) = Ae^{kt}.$

$y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945$

$\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\ ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}$

(b)
$Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}$