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2 years ago

6

Translate to an algebraic expression. The sum of a and n. The translation is

2 answers:

Leno4ka [110]2 years ago

6 0

Bro stop doing math u don’t it won’t help u I’m real life.

aksik [14]2 years ago

5 0

The translation of the sum a and n is a+n

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schepotkina [342]

Answer:

3.56 cm.

Step-by-step explanation:

1/3 π r^2 h = V

1/3 π (5.13)^2 h = 98

h = 98 / (1/3 π (5.13)^2)

= 98 / 27.55899

= 3.556.

3 0

2 years ago

Emily and Kylie are 140 miles apart, traveling towards each other. If Kylie travels 12 mph and Emily travels 8 mph how long unti

Bogdan [553]

Answer: 7 hr

Step-by-step explanation:

Given

Distance between Emily and Kylie is $140\ \text{miles}$

Speed of Emily $v_e=12\ mph$

Speed of Kylie $v_k=8\ mph$

Distance traveled by Emily $\Rightarrow d_e=12t$

Distance traveled by Kylie $\Rightarrow d_k=8t$

$\Rightarrow 12t+8t=140\\\Rightarrow 20t=140\\\Rightarrow t=7\ hr$

They will meet after 7 hour.

3 0

2 years ago

drek231 [11]

Umm is this the problem we need to solve?

6 0

3 years ago

Which choice best describes the following sentence?

Wittaler [7]

Answer:

The choice that best describes the given sentence above is, it is a complete and correct sentence. What makes this sentence complete is having both the subject and the verb in a simple form, and still expresses a complete thought. The verb and simple predicate "volunteered" is enough to describe the subject "Cassidy".

6 0

3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago