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Archy [21]
3 years ago
6

A rectangle has length 7 inches and width 3 inches.

Mathematics
1 answer:
Naya [18.7K]3 years ago
6 0

Answer:

The area of rectangle is 21 in^2

Step-by-step explanation:

we know that

The area of rectangle is equal to

A=LW

where

L is the length

W is the width

In this problem we have

L=7\ in

W=3\ in

substitute

A=(7)(3)=21\ in^2

The draw in the attached figure

therefore

The area of rectangle is 21 in^2

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Write an equation in point-slope form of the line that passes through the points
Aleksandr-060686 [28]

Hello!

Step-by-step explanation:

Slope: \frac{Y^2-Y^1}{X^2-X^1}=\frac{rise}{run}

\frac{(-2)-(-4)=-2+4=2}{3-2=1}=\frac{2}{1}=2

<u><em>Answer: 2</em></u>

<u><em>Slope is 2.</em></u>

Hope this helps!

Thanks!

-Charlie

Have a great day!

:)

:D

4 0
3 years ago
6. a. Sixty students in a class took an examination in Physics and Mathematics. If 17 of them passed Physics only, 25 passed in
Ivahew [28]

Let C be the set of all students in the <u>c</u>lassroom.

Let P and M be the sets of students that pass <u>p</u>hysics and <u>m</u>ath, respectively.

We're given

n(C) = 60

n(P \cap M') = 17

n(P \cap M) = 25

n((P \cup M)') = n(P' \cap M') = 9

i. We can split up P into subsets of students that pass both physics and math (P\cap M) and those that pass only physics (P\cap M'). These sets are disjoint, so

n(P) = n(P\cap M) + n(P\cap M') = 25 + 17 = \boxed{42}

ii. 9 students fails both subjects, so we find

n(C) = n(P\cup M) + n(P\cup M)' \implies n(P\cup M) = 60 - 9 = 51

By the inclusion/exclusion principle,

n(P\cup M) = n(P) + n(M) - n(P\cap M)

Using the result from part (i), we have

n(M) = 51 - 42 + 25 = 34

and so the probability of selecting a student from this set is

\mathrm{Pr}(M) = \dfrac{34}{60} = \boxed{\dfrac{17}{30}}

7 0
2 years ago
Yesterday, Caitlin swam 20 laps in the pool. Hannah swam 4/5 as many laps As Caitlin. How many laps did Hannah swim?
Misha Larkins [42]

Caitlin = 20 laps

Hannah = 4/5 as many laps

Multiply the laps swan by Caitlin (20) by 4/5

20 x 4/5 = 80/5 = 16

Hannah swam 16 laps

6 0
1 year ago
Find the inverse function of
shtirl [24]

Answer:

f(x) = 20x - 4 \\ substitute \: y \: for \: f(x) \\ y = 20x - 4 \\ interchange \: x \: and \: y \\ x = 20y - 4 \\ swap \: the \: sides \: of \: the \: equation \\ 20y - 4 = x \\ move \: the \: constant \: to \: the \: right \: hand \\ 20y = x + 4  \\ divide \: both \: sides \: by \: 20 \\ y =  \frac{1}{20} x +  \frac{1}{5}  \\ substitute \: f {}^{ - 1} (x) \: for \: y \\ f {}^{ - 1} (x) =  \frac{1}{20} x +  \frac{1}{5}

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

Thus, domain of f(x): x∈R = range of f¯¹(x)

and range of f(x): x∈R =domain of f¯¹(x)

3 0
2 years ago
Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat
yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

P(t) = 0.8t^{2} + 6t + 19000

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

P(t) = 0.8t^{2} + 6t + 19000

0.8t^{2} + 6t + 19000 = 34200

0.8t^{2} + 6t - 15200 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

0.8t^{2} + 6t - 15200 = 0

So a = 0.8, b = 6, c = -15200

Then

\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356

t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14

t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0
3 years ago
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