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3 years ago

What is -4/9 x -5/8 this is for math

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

5 0

0.2778 this is the answer for your question

Whitepunk [10]3 years ago

5 0

$\frac{-4}{9} * \frac{-5}{8}\\$
The negatives cancel out;
$\frac{20}{72} \\
\frac{5}{18} \ or \ .277778...$

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What is another way to write the expression t⋅(14−5) ? (t⋅14)⋅5 t⋅14−5 (t⋅14)−5 (t⋅14)−(t⋅5)

Finger [1]

Another way to write the expression the answer is this (t.14)-(t.5)

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3 years ago

Read 2 more answers

the line joining A(a, 3) to B(2 -3) is perpendicular to the line joining C(10,1) to B. The value of a is?

RideAnS [48]

Well, first off, let's find what is the slope of BC

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 2}}\quad ,&{{ -3}})\quad 
% (c,d)
C&({{ 10}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-(-3)}{10-2}\implies \cfrac{1+3}{10-2}
\\\\\\
\cfrac{4}{8}\implies \cfrac{1}{2}$

now, a line perpendicular to that one, will have a negative reciprocal slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{2}\\\\
slope=\cfrac{1}{{{ 2}}}\qquad negative\implies -\cfrac{1}{{{ 2}}}\qquad reciprocal\implies - \cfrac{{{ 2}}}{1}\implies \boxed{-2}$

now, we know the slope "m" of AB is -2 then, thus

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ a}}\quad ,&{{ 3}})\quad 
% (c,d)
B&({{ 2}}\quad ,&{{ -3}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-3-3}{2-a}=\boxed{-2}
\\\\\\
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\\\\\\
-1=a$

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3 years ago

Factor out the coefficient of the variable term The expression 1/10k-7/10 factored is

Goshia [24]

Answer:

The answer: 1/10x (k-7)

Step-by-step explanation:

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3 years ago

Will give Brainliest if you get it right

fenix001 [56]

Answer:

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Step-by-step explanation:

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sqrt(8^2 + 7^2) =

sqrt(64 + 49) =

sqrt( 113) =

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3 years ago

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Yuri [45]

Answer:

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Step-by-step explanation:

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