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2 years ago

Which equation can not be used to find the area of a triangle

Mathematics

1 answer:

maxonik [38]2 years ago

7 0

Answer:

Step-by-step explanation:

because triangle don't work that way

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3a. The graph shows a school's profit P for selling x lunches on one day. PART A: What is the school's profit for selling 120 lu

konstantin123 [22]

Answer:

where is the graph

Step-by-step explanation:

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3 years ago

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What is the answer to -7=-12 (-2)+B please help!

insens350 [35]

I would solve it for B, but that's just what I think. So you would do -12 time -2 which is 24, then subtract 24 from both sides so that you would get B=-31

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3 years ago

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Pls help will give brainliest

Shtirlitz [24]

Answer:

C, C, C

Hope This Helps! Have A Nice Day!!

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3 years ago

Find the volume of a right circular cone that has a height of 16.8 in and a base with a

alexgriva [62]

Answer:

Volume of right circular cone is 388.43 in³

Step-by-step explanation:

Height of circular cone = 16.8 in

Radius of circular cone = 4.7 in

We need to find Volume of right circular cone

The formula used for calculating volume of a right circular cone is: $Volume=\pi r^2\frac{h}{3}$

Putting values and finding volume

$Volume=\pi r^2\frac{h}{3}\\Volume=3.14\times (4.7)^2 \times \frac{16.8}{3}\\Volume=3.14\times 22.09 \times 5.6\\Volume= 388.43 \:in^3$

So, Volume of right circular cone is 388.43 in³

5 0

2 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