Find the value of x to the nearest tenth.

Mathematics

2 answers:

Viefleur [7K]3 years ago

7 0

Answer:

x = 7.8

Step-by-step explanation:

a^2 + b^2 = c^2

4^2 + 2^2 = c^2

16 + 4 = c^2

c^2 = 20

c = 4.5

a^2 + b^2 = c^2

a^2 + 4.5^2 = 9^2

a^2 + 20.25 = 81

a^2 = 81 - 20.25

a^2 = 60.75

a = 7.8

Digiron [165]3 years ago

7 0

Answer:

Step-by-step explanation:

Using Pythagoreus Theorem

In the smaller rectangle

2²+4² = h²

20 = h²

h = √20

h = 2√5

Therefore two sides of the rectangle will 2√5

No using pythagoreus theorem in the bigger triangle

x² + (2√5)² = 9²

x² = 9² - (2√5)²

x² = 81 - 20

x = √61

x = 7.8102496759

When rounded off to nearest tenth the hundreth digit is in 0-5 range

x = 7.81

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What is the solution(s) to the system of equations y = x² +4 and y = 2x + 4

bija089 [108]

Answer:

$(x,y) = (0,4)~ \text{and}~ (x,y) = (2,8)$

Step by step explanation:

$y = x^2 +4~~~~~~~~~...(i)\\\\y = 2x +4~~~~~~~~~...(ii)\\\\\text{From (i) and (ii):}\\\\~~~~~~x^2 +4 = 2x+4\\\\\implies x^2 = 2x\\\\\implies x^2 -2x = 0\\\\\implies x(x-2) = 0\\\\\implies x =0,~ x = 2$

$\text{Substitute x = 0 in eq (i):}\\\\y = 0^2 +4 = 4\\\\\text{Substitute x = 2 in eq (i):}\\\\y=2^2+4 = 4+4 = 8\\\\\text{Hence,}~ (x,y)= \{(2,8), (0,4)\}$