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2 years ago

Solve for t: d=rt. I need some help.

Mathematics

2 answers:

Elena L [17]2 years ago

7 0

D = rt
/r /r

D/r = T

*sample text*

Greeley [361]2 years ago

7 0

Dr=t is what I think the answer should be

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A 28 feet support beam is cut into two pieces so that one is six times the length of the other. How long is the longest piece

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The longest piece is 24 ft and the smallest piece is 4 ft

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Tell whether (2,0) is a solution of y=7x+ 2.

sattari [20]

Answer:

Step-by-step explanation:

7x2=14, 14+2=16, 16 is not 0.

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In a city school, 70% of students have blue eyes, 45% have dark hair, and 30% have blue eyes and dark hair. What is the probabil

Travka [436]

Answer:

The correct answer option is 43%.

Step-by-step explanation:

We are given that 70% of students have blue eyes, 45% have dark hair, and 30% have blue eyes and dark hair.

We are to find the probability of a student getting selected will have dark hair with blue eyes.

P(D|B) = P(D∩B) / P(B)

Substituting the given values in it to get:

P(D|B) = 0.3 / 0.7 = 0.428

Rounding it to the nearest whole percent we get:

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4 0

2 years ago

Solve the equation. 3 = n + 4 Question 3 options: 7 1 -1 12

zysi [14]

Hey!

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Steps To Solve:

~Subtract n to both sides

3 - n = n + 4 - n

~Simplify

3 - n = 4

~Subtract 3 to both sides

3 - n - 3 = 3 - 4

~Simplify

n = -1

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Answer:

$\large\boxed{n~=~-1}$

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Hope This Helped! Good Luck!

6 0

2 years ago

Situation:

Gala2k [10]

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

Given formula:

$\sf N=N_0e^{-kt}$

$\sf N_0$ = initial mass (at time t=0)
N = mass (at time t)
k = a positive constant
t = time (in days)

Given values:

$\sf N_0$ = 11 g
k = 0.125

Half-life: The time required for a quantity to reduce to half of its initial value.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

$\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}$

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

brainly.com/question/28016999

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1 year ago

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