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3 years ago

FOR 15 POINTS Which is the solution to the equation 0.5 x + 4.2 = 5.9? Round to the nearest tenth if necessary. 0.9 3.4 5.1 20.2

Mathematics

2 answers:

alina1380 [7]3 years ago

7 0

Answer:

3.4

Step-by-step explanation:

weqwewe [10]3 years ago

3 0

Answer:

3.4

Step-by-step explanation:

0.5x+4.2=5.9

Subtract 4.2 from both sides:

0.5x=1.7

Divide both sides by 0.5:

x=3.4

Hope this helps!

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Solve- cbb to work it out

zloy xaker [14]

<h3>Answers: </h3>

➝ Hypotenuse of triangle ( a ) = 21.63 mm

➝ Hypotenuse of triangle ( b ) = 150 mm

➝ Hypotenuse of triangle ( c ) = 111.80 mm

$\quad\rule{300pt}{1.5pt}\quad$

<h3>Solution:</h3>

We have to find the length of hypotenuse in the given 3 triangles, which can be done by using Pythagoras theorem.

Pythagoras theorem states that :

" In a right angled triangle, the square of hypotenuse side is equal to the sum of square of other two sides "

$\qquad \bull \:{\pmb{\mathfrak{ h^2 = b^2 + p^2}}}$

And we have to convert the answer to the units indicated in red i.e, in mm.

Since 1cm = 10 mm, we will convert the given values of length of side in mm before putting the values in the formula

For triangle ( a )

$:\implies\qquad \sf{ h^2 = b^2 + p^2}$

$:\implies\qquad \sf{ h =\sqrt{b^2 + p^2}}$

$:\implies\qquad \sf{h=\sqrt{ (12)^2 + (18)^2 }}$

$:\implies\qquad \sf{ h= \sqrt{144 + 324}}$

$:\implies\qquad \sf{ h = \sqrt{468}}$

$:\implies\qquad\underline{\underline{\pmb{ \sf{ h = 21.63 mm}}}}$

For triangle ( b )

$:\implies\qquad \sf{ h^2 = b^2 + p^2}$

$:\implies\qquad \sf{h =\sqrt{b^2 + p^2} }$

$:\implies\qquad \sf{ h = \sqrt{(90)^2+(120)^2}}$

$:\implies\qquad \sf{ h=\sqrt{8100+14400}}$

$:\implies\qquad \sf{ h =\sqrt{22500}}$

$:\implies\qquad\underline{\underline{\pmb{ \sf{h = 150mm}}} }$

For triangle ( c )

$:\implies\qquad \sf{h^2 = b^2 + p^2 }$

$:\implies\qquad \sf{ h=\sqrt{b^2 + p^2}}$

$:\implies\qquad \sf{ h =\sqrt{(100)^2)+(50)^2}}$

$:\implies\qquad \sf{ h=\sqrt{10000+2500}}$

$:\implies\qquad \sf{ h =\sqrt{12500}}$

$:\implies\qquad \underline{\underline{\pmb{\sf{h = 111.80mm}}} }$

4 0

2 years ago