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What is the surface area of this figure

Alenkinab [10]

Answer:

10in+10in+15in=35 inch

4 0

3 years ago

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find the equation of a line that is perpendicular to line g that contains (P,Q) will give brainlest to whoever answers

Fudgin [204]

Answer:

Consider points (-3,6) and (0,5)

$gradient \: \: m = \frac{(6 - 5)}{( - 3 - 0)} \\ m = - \frac{1}{3} \\ for \: perpendicular \: gradient \: m {}^{.} \\ m {}^{.} \times m = - 1 \\ - \frac{1}{3} m {}^{.} = - 1 \\ m {}^{.} = 3 \\ from \: y = mx + c \\ at \:(0,5) \: \: 5 = 0 \times 3 + c \\ c = 5 \\ hence \: equation \: is \: \: y = 3x + 5$

3 0

3 years ago

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A starting balance of $1,028.23, during the month there were checks in written for $300, $279.50 and $459.71 and a deposit of $1

madreJ [45]

Answer:

overdraft. A check written for more money than your account contains. floating a check. Writing a check hoping to deposit money to cover the check before it clears the bank (very risky) checkbook register.

Step-by-step explanation:

overdraft. A check written for more money than your account contains. floating a check. Writing a check hoping to deposit money to cover the check before it clears the bank (very risky) checkbook register.

4 0

3 years ago

Situation: a 25 gram sample of a substance that’s used for drug research has a K-value of 0.1205

Lelu [443]

The decay model is given as

$N=N_0e^{-Kt}$

The substance's half-life is the time it takes for the substance to decay to half its original amount. In other words, it's the time $t$ such that $N=\dfrac{N_0}2$ . Substituting into the model, you have

$\dfrac12=e^{-Kt}$

You can solve for $t$ now.

$\dfrac12=e^{-Kt}$
$\implies\ln\dfrac12=\ln e^{-Kt}$
$\implies-\ln2=-Kt\ln e$
$\implies\dfrac{\ln2}K=t\approx5.7523\text{ days}$

8 0

3 years ago

Combining like terms although there has to be two terms<br> y-9+y-4

liberstina [14]

Answer:

2y - 13

Step-by-step explanation:

y - 9 + y - 4

2y - 9 - 4

2y - 13

3 0

3 years ago

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Please help me PLEASE