(03.03 LC)

<img src="https://tex.z-dn.net/?f=y%20-%20xy%20%2B%20x%20-%201" id="TexFormula1" title="y - xy + x - 1" alt="y - xy + x - 1" ali

nata0808 [166]

I believe it should be =(1-x) multiple by (y-1)
(1-x) x (y-1)

8 0

2 years ago

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Part 3: The Space Inside! Find the volume of the shipping box using the two methods and show your work: Packing cubes Using the

kirill115 [55]

Answer:shipping box=3.75*3*3.25=36.562 cubic feet.in width we have 15 cubes,in the legnt 12 while in height 13.with a total of 2340 cubes

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

6 0

2 years ago

Hi, I am taking PSAT and I am working on the math section. I need help with answering this math type of question. Please give th

ValentinkaMS [17]

By using the concept of uniform rectilinear motion, the distance surplus of the average race car is equal to 3 / 4 miles. (Right choice: A)

<h3>How many more distance does the average race car travels than the average consumer car?</h3>

In accordance with the statement, both the average consumer car and the average race car travel at constant speed (v), in miles per hour. The distance traveled by the vehicle (s), in miles, is equal to the product of the speed and time (t), in hours. The distance surplus (s'), in miles, done by the average race car is determined by the following expression:

s' = (v' - v) · t

Where:

v' - Speed of the average race car, in miles per hour.
v - Speed of the average consumer car, in miles per hour.
t - Time, in hours.

Please notice that a hour equal 3600 seconds. If we know that v' = 210 mi / h, v = 120 mi / h and t = 30 / 3600 h, then the distance surplus of the average race car is:

s' = (210 - 120) · (30 / 3600)

s' = 3 / 4 mi

The distance surplus of the average race car is equal to 3 / 4 miles.

To learn more on uniform rectilinear motion: brainly.com/question/10153269

#SPJ1

4 0

1 year ago

Factoring polynomials

-Dominant- [34]

1. x(2x^2-x+4)
2.-3(y-1)(y+2)
3.-(11w^2+18w-1)

7 0

3 years ago