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Rus_ich [418]
3 years ago
9

Find the area of the triangle.

Mathematics
2 answers:
Butoxors [25]3 years ago
6 0

Answer:

56

Step-by-step explanation:

Sana po makatulog #carry on learning

Aneli [31]3 years ago
6 0

Answer:

28

Step-by-step explanation:

area of triangle is 1/2xbxh

so assuming 1/2x8x7

so it is 28 sq.ft

hope it helps you

please mark me as brainlist

You might be interested in
2 The value of y is directly proportional
rusak2 [61]

Answer:

y = 17.5

Step-by-step explanation:

Given that y is directly proportional to x then the equation relating them is

y = kx ← k is the constant of proportion

To find k use the condition y = 35 when x = 140, then

35 = 140k ( divide both sides by 140 )

0.25 = k

y = 0.25x ← equation of proportion

When x = 70, then

y = 0.25 × 70 = 17.5

4 0
3 years ago
A new employee charged $3980 on his credit card to relocate for his first job.After noticing that the interest rate for his bala
olga nikolaevna [1]

Answer:

that should be how you would solve those questions..

8 0
3 years ago
Write the sum as a product simplify the product-2 + -2 + -2 + -2 equals
yuradex [85]
-2+-2+-2+-2
-4+-4
=-8
hope this helps:)
6 0
3 years ago
Solve this .....................​
ycow [4]

Answer:

Step-by-step explanation:  3root5x^2 + 25x - 10root5 = 0

3xroot5 + 25x - 10root5 = 0      [ root x^2 = x]

28x root5 = 10 root5         [ -10root5 turns to 10 root5 when transferred to RHS]

28x root 5/root5 =10

28x=10

x = 10/28

x = 0.35

Hope it helped u,

pls mark as the brainliest

^v^

5 0
3 years ago
PLEASE HELP!!! Find the equation , in the standard form of the line passing through the points (3,-4) and (5,1)
ExtremeBDS [4]
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 3 &,& -4~) 
%  (c,d)
&&(~ 5 &,& 1~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}

\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby 
\begin{array}{llll}
\textit{now let's multiply both}\\
\textit{sides by }\stackrel{LCD}{2}
\end{array}
\\\\\\
2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23}
\\\\\\
\textit{and if we multiply both sides by -1}\qquad 5x-2y=23

side note:  multiplying by the LCD of both sides is just to get rid of the denominators
5 0
3 years ago
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