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4 years ago

Simplify: (8)2 (−1)3 (2)5

Mathematics

2 answers:

Elis [28]4 years ago

8 0

The answer is -480 hope this helps

kykrilka [37]4 years ago

4 0

I think the answer is -480.

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Find the area of a kite with diagonals 15 m and 7 m.

lozanna [386]

Answer:

Step-by-step explanation:

The formula of an area of a kite:

$A=\dfrac{e\cdot f}{2}$

<em>e, f</em><em> - diagonals</em>

<em />

We have <em>e = 15m, f = 7m.</em>

Substitute:

$A=\dfraC{(15)(7)}{2}=\dfrac{105}{2}=52.5\ m^2$

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3 years ago

You had $22 to spend on four raffle tickets. After buying them you had $6. How much did each raffle ticket cost?

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22-6=16. 16÷4=4. Each ticket costs $4.

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4 years ago

At a football stadium, 4% of the fans in attendance were teenagers. If there were 120 teenagers at the football stadium, what wa

GuDViN [60]

Answer:

3000 people

Step-by-step explanation:

4% teenagers --> 120 teenagers were there

120/4 means 1% so 30 is 1%

1% times 100 is 100% so 30 times 100 equals 100% so 3000 is 100% of the people at the stadium

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3 years ago

Read 2 more answers

The sum of the lengths of two opposite sides of the circumscribed quadrilateral is 12 cm, the length of a radius of the circle i

nasty-shy [4]

Answer:

60cm^2

Step-by-step explanation:

We assume that is a circumscribing quadrilateral, rather than one that is circumscribed. It is also called a "tangential quadrilateral" and its area is ...

K = sr

where s is the semi-perimeter, the sum of opposite sides, and r is the radius of the incircle.

K = (12 cm) (5cm) = 60 cm²

_____

A quadrilateral can only be tangential if pairs of opposite sides add to the same length. Hence the given sum is the semiperimeter.

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3 years ago

The population of a certain species of insect is given by a differentiable function P, where P(t) is the number of insects in th

Step2247 [10]

Answer:

$\frac{dP}{dt} \ =\ \frac{1}{5}P$

Step-by-step explanation:

Given

Variation: Directly Proportional

Insects = 10 million when Rate = 2 million

i.e. $P = 10\ million$ when $\frac{dP}{dt} = 2\ million$

Required

Determine the differential equation for the scenario

The variation can be represented as:

$\frac{dP}{dt} \ \alpha\ P$

Which means that the rate at which the insects increase with time is directly proportional to the number of insects

Convert to equation

$\frac{dP}{dt} \ =\ k * P$

$\frac{dP}{dt} \ =\ kP$

Substitute values for $\frac{dP}{dt}$ and P

$2\ million = k * 10\ million$

Make k the subject

$k = \frac{2\ million}{10\ million}$

$k = \frac{2}{10}$

$k = \frac{1}{5}$

Substitute $\frac{1}{5}$ for k in $\frac{dP}{dt} \ =\ k * P$

$\frac{dP}{dt} \ =\ \frac{1}{5}* P$

$\frac{dP}{dt} \ =\ \frac{1}{5}P$

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3 years ago