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3 years ago

List of particle products of 40 × 28

Mathematics

2 answers:

NARA [144]3 years ago

8 0

40x28 just multiply them and you get
1,120

katrin [286]3 years ago

5 0

40x28

First take 0x8=0
Then take 40*8=320
Next, 20*0=0
Then, 40*20=800
Finally, add them all together to get 1120!

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A farmer sells four of his farm products Maize, Potatoes, carrots and tomatoes in each of 2 towns into classes of 3 customers. C

nordsb [41]

Answer:

Step-by-step explanation:

7 0

3 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

3 years ago

-3(5)^2-(-2)(-8) simplify

avanturin [10]

To do this, we're going to use the order of operations (PEMDAS):

P - Parentheses

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

First let's do parentheses, there isn't anythig in parentheses we need to simplify, so we can skip this step.

Next let's look for exponents. I see we have a $5^2$ so let's replace that with $5^2=25$ :

$-3(25)-(-2)(-8)$

Now let's look for multiplcation. We know that things that are right next to eachother in parentheses represent multiplcation, so let's simply this more:

$-3(25)=-75$

$(-2)(-8)=16$

$-75-16$

And now we're left with a simple problem we know how to solve.

Answer: $-91$

Hope this helps!

3 0

3 years ago

GET POINTS look at pictur 15 POINTS

kap26 [50]

2. y=2x+3

When x =0 and y= something, the 'something' is your y-intercept which is b in the y=mx+b formula. Then we look at what is changed in the x and y value to determine the m part. the x increase by 1 and y increase by 2 so the m part is 2.

3. y=-7x
Same as the last one, there is no y-inter this time because the y=0 when x=0/ So the y deceased by 7 each time and x incease by 1 so it is -7 for the m part.

6 0

3 years ago

Read 2 more answers

How do you know whether you have to write a sum in simplest form

skelet666 [1.2K]

I do it when the numbers are to big but that's me

5 0

3 years ago