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AnnZ [28]
3 years ago
8

Write an expression that uses partial products to multiply 8× 64

Mathematics
2 answers:
Radda [10]3 years ago
7 0
8x64 equals one thirty two
g100num [7]3 years ago
4 0
8 x 64=
8 x 32 x 2
64 x 4 x 2
128 x 2 x 2
=512

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Kol and Kara solve this problem in two different ways the perimeter of a rectangle is 46 centimeters it’s length is 8 centimeter
JulsSmile [24]

Answer:

15cm

Step-by-step explanation:

because length is given as 8 cm and perimeter is length+width+length+width

so if you take 8cmX2 since its taken twice to get the perimeter and 15cmX2 you get 46cm which originally the question said was the peemiter

5 0
2 years ago
Read 2 more answers
Clare wants to mail a package that weighs 4 1/2 ponds what could be its volume in liters
Sati [7]

Answer:

The volume in liters is 2.041165665 liter.

Step-by-step explanation:

Given : Clare wants to mail a package that weighs 4\frac{1}{2} pounds.

To find : What could be its volume in liters ?

Solution :

We know that,

Pound (lb) is a unit of Weight used in Standard system.

Liter (l) is a unit of Volume used in Metric system.

To convert 1 pound into liter is

1 pound (lb) = 0.45359237 liter (l)

4\frac{1}{2} pounds in simpler fraction is \frac{9}{2}

Converting into liter,

\frac{9}{2} pound (lb)=\frac{9}{2}\times 0.45359237 liter (l)

\frac{9}{2} pound (lb) = 2.041165665 liter (l).

Therefore, the volume in liters is 2.041165665 liter.

7 0
3 years ago
Solve by substitution x-y=-11 y+7=-2x
Bess [88]

x - y = - 11   .............( 1 )

y + 7 = - 2x ...............( 2 )

from equation ( 1 )

x - y = - 11

x   = -11 + y  ...........( 3)      

putting x in equation  ( 2 )

y + 7 = - 2 x  

y  + 7 = -2 ( -11 + y )             y + 7 = 22 - 2 y                       y + 2 y =22 - 7            

3 y = 15                  y = 15 / 3

putting value of y in equation 3

x   = -11+ ( 13 /5 )                       x = -33 /3 + 15 / 3         ( l.c.m)

x  = -17 / 5                      

check      

x = -11 + y

- 17 / 5 = -11 + 13 / 5

-17 /5   = -17 / 5

4 0
3 years ago
Can anyone help! I have done 1 and 3 but I’m lost on the rest! Will give extra points!
MrRissso [65]

Answer:

i think thats good but for me umm so good hehe

4 0
2 years ago
Read 2 more answers
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
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