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ipn [44]
3 years ago
11

help meeeeeeee please Nо files оr Im sоrry I dоnt understand this I will mark you brainlist if you answer all of them!!!! PLEASE

HELPPPP

Mathematics
1 answer:
Yakvenalex [24]3 years ago
7 0

Lesson 4: Different equations can describe the situation because the equation doesn't have to be the same equation. you can have 21 - 11 and get 10, but have 2 + 8 and get 10.

Lesson 1: A tape diagram is used to represent a situation by having numbers on each lets say stamp you could have 20 as a product and have 4 stamps with 5 on each of them.

I don't know how to make a tape diagram on a computer.

Lesson 2; What it means that an equation is either true or false is if you get the correct answer to the equation.

Lesson 5: A fraction is made out of fractions or decimals because there is always an equal fraction. and you can represent a fraction like 10/10 as 1.00 as a decimal or 1/10 as 0.1

Lesson 3: A balance hanger will give you steps to find an unknown amount in an associated equation if the left side of the hanger has 5 squares on it and the number of 15 you would know that each square is equal to 3. and a true equation would tell you what you would add or multiply or divide or subtract.

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c

Step-by-step explanation:

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In ∆PQR, p=13 inches, q=18 inches and r= 12 inches. Find the area of ∆PQR to the nearest square inch.
sveta [45]

Given data:

The first side of the triangle is p=13 inches.

The second side of the triangle is q=18 inches.

The third side of the triangle is r= 12 inches.

The semi-perimeter is,

\begin{gathered} s=\frac{p+q+r}{2} \\ =\frac{13\text{ in+18 in+12 in}}{2} \\ =21.5\text{ in} \end{gathered}

The expression for the area of the triangle is,

\begin{gathered} A=\sqrt[]{s(s-p)(s-q)(s-r)_{}} \\ =\sqrt[]{21.5\text{ in(21.5 in-13 in)(21.5 in-18 in)(21.5 in-12 in)}} \\ =\sqrt[]{(21.5\text{ in)(8.5 in)(3.5 in)(9.5 in)}} \\ =77.95in^2 \end{gathered}

Thus, the area of the given triangle is 77.95 sq-inches.

7 0
1 year ago
A parcel delivery service will deliver a package only if the length plus girth (distance around) does not exceed 84 in. What are
Lesechka [4]

Answer:

a) \frac{28}{\pi} in

b) 28 in

c) 784 in²

Step-by-step explanation:

Let the length be 'L'

and the radius be 'r'

Thus, according to the question

L + 2πr = 84 in

L = 84 - 2πr   ............(1)

Volume of the cylinder, V = πr²L

substituting the value of L from 1, we get

V =  πr²(84 - 2πr)

or

V = 84πr² - 2π²r³

for points of maxima, differentiating the above equation and equating it to zero

\frac{dV}{dr}=\frac{d(84\pi r^2-2\pi^2 r^3))}{dr}

or

2(84)πr - 3(2)π²r² = 0

or

2πr(84 - 3πr) = 0

or

r = 0    and    84 - 3πr = 0

or

⇒ 3πr = 84

or

⇒ r = \frac{28}{\pi} in

since, the radius cannot be zero therefore, r = 0 is neglected

Therefore,

a) The radius of the largest cylindrical package = \frac{28}{\pi} in

b) from  (2)

L = 84 - 2πr

or

⇒ L = 84 - 2\pi\times\frac{28}{\pi}

or

⇒ L = 84 - 56 = 28 in

The length of the largest cylindrical package = 28 in

c ) The volume of the largest cylindrical package ,V = πr²L

= \pi\times\frac{28}{\pi}\times28

= 784 in²

7 0
3 years ago
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