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Marysya12 [62]
3 years ago
5

What is the highest common factor of 65 and 56?

Mathematics
2 answers:
Evgesh-ka [11]3 years ago
6 0
Highest common factor of 56&65 is 1
gizmo_the_mogwai [7]3 years ago
5 0

Step-by-step explanation:

don't you know how to do HCF

look for a number that can divide all the numbers

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Calculate the average for this list of numbers 1.90 7.99 4.33 5.21
malfutka [58]
What you do is add up all the values, then divide by the number of values.
\frac{1.90+7.99+4.33+5.21}{4}=\frac{19.43}{4}=4.8575
The average of the data is 4.8575.
8 0
3 years ago
HELPPPP HELP HELP HELP HELPPPPPPPPPP
eduard

Answer:

x = -7

Step-by-step explanation:

Use the distibutive property on both side:

Right side:

4 ( 1 - x )

( 4 x 1 ) + 4 x -X )

4 - 4x

Left side:

-3 ( x + 1 )

( -3 x X ) + ( -3 x 1 )

-3x -3

4 - 4x + 2x = -3x - 3

Combine like terms:

( 2x + ( -4x ) ) + 4 = -3x - 3

-2x + 4 = -3x - 3

Add 3 to each side:

-2x + 7 = -3x

add 2x to each side:

7 = -x

Divide each side by -1:

-7 = x

8 0
3 years ago
a school is running a fundraiser. For every $75 worth of wrapping paper sold, the school receives $20. How much wrapping paper m
Reptile [31]

Answer:

125

Step-by-step explanation:

if the school gets $20 per wrapping paper sold

then in order to earn 2500 the school must sell 2500/20=125

5 0
3 years ago
Read 2 more answers
Verify tan(α+β) = [tan(α)+tan(β)] / [1 + tan(α)tan(β)]
mr Goodwill [35]

Answer:

Proved

Step-by-step explanation:

To Prove: tan(\alpha+\beta) =\dfrac{tan(\alpha)+tan(\beta)}{1 + tan(\alpha)tan(\beta)}

Proof:

Now: tan \theta =\dfrac{sin\theta }{cos \theta}

Therefore:

tan (\alpha+\beta)=\dfrac{ sin (\alpha+\beta)}{cos (\alpha+\beta) }

Applying these angle sum formula

sin (\alpha+\beta)=sin \alpha cos \alpha + sin \beta cos \beta\\cos (\alpha+\beta)=cos \alpha cos \beta - sin \alpha sin \beta

tan (\alpha+\beta)=\dfrac{ sin \alpha cos \alpha + sin \beta cos \beta}{cos \alpha cos \beta - sin \alpha sin \beta }

Divide all through by cos \alpha cos \beta

tan (\alpha+\beta)=\dfrac{ (sin \alpha cos \alpha)/(cos \alpha cos \beta) + (sin \beta cos \beta)/(cos \alpha cos \beta)}{(cos \alpha cos \beta)/(cos \alpha cos \beta) - (sin \alpha sin \beta)/(cos \alpha cos \beta) }\\\\tan (\alpha+\beta)=\dfrac{\frac{sin \alpha}{cos \alpha}+\frac{sin \beta}{cos \beta} }{1-tan \alpha tan \beta} \\$Therefore:\\\\tan (\alpha+\beta)=\dfrac{tan \alpha+tan \beta}{1-tan \alpha tan \beta}

=sin \alpha cos \beta + cos \alpha sin \beta/cos \alpha cos \beta/cos \alpha cos \beta- sin \alpha sin \beta/cos \alpha cos \beta  

=sin \alpha/cos \alpha + sin \beta/cos \beta/1-tan \alpha tan \beta  

=tan A + tan B/1-tan A tan B

5 0
4 years ago
Pls help me :(<br><br>Give four possible solutions for x in the inequality x &lt; -4.​
Margaret [11]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Since 'x' is less than -4, it can just be any value below -4.

Here are some examples:

x = -6, x = -9, x = -10 and x = -5

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

5 0
3 years ago
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