All categories

3 years ago

What is the range of possible sizes for side x

Mathematics

2 answers:

Gnesinka [82]3 years ago

8 0

2.8 easy just think about it

Molodets [167]3 years ago

4 0

Answer: the real answer is 0<x<5.6

Step-by-step explanation:

You might be interested in

How do i solve 9/10 y = -36

katrin2010 [14]

Answer:

-40

Step-by-step explanation:

Multiply both sides by the reciprocal of 9/10 to get rid of the fraction

$\frac{10}{9}$ × $\frac{9}{10}$ ×y=-36 × $\frac{10}{9}$

y=-36× $\frac{10}{9}$

y= $\frac{-360}{9}$

y=-40

3 0

3 years ago

6. What value of a makes this equation true?

arlik [135]

The answer I believe is c

8 0

3 years ago

If X^2= 3+2√2 what is the value of √x+1/√x

Nikolay [14]

Step-by-step explanation:

Let √x - 1/√x = a

Squaring both the sides,

x + 1/x - 2 = a^2

Putting the value,

3–2√2 + 1/(3–2√2) - 2 = a^2

a^2 = 1 - 2√2 + 1/(3–2√2)

= [(3–2√2) (1–2√2) + 1] / 3–2√2

= {3 - 8√2 + 9} / 3–2√2

= [12 - 8√2] / 3–2√2

Rationalising both the sides

= {(12 - 8√2)(3+2√2)} ÷ (9–8)

= 36 + 24√2 - 24√2 + 16(2)

= 36 - 32

=> 4

a^2 = 4

a = √4

a = 2, -2

So,

√x - 1/√x = a = 2, -2.

5 0

3 years ago

Please help!! 20 points

Marianna [84]

If the width is w then the length = 2w + 40, so the area

= w(2w + 40)

= 2w^2 + 40 w

6 0

4 years ago

Read 2 more answers

How do i solve this question, anyone help

Degger [83]

Answer:

a. $\sqrt{52}$ cm or 7.21cm

b. $\sqrt{170}$ cm or 13.04cm

c. x= $\sqrt{136}$ cm or 11.66cm, y= $\sqrt{191}$ cm or 13.82cm

Step-by-step explanation:

a. You have to find the length of the other two sides of the triangle using the information already given. The first side is 6cm and the other is 12-9=4cm. Because it's a right-angled triangle you can use pythagoras

$c^2=a^2+b^2\\x^2=6^2+4^2\\x^2=36+16\\x^2=52\\x=\sqrt{52}cm \\x=7.21cm$

b. You can use pythagoras again because it's a right-angle triangle

$x^{2} =7^2+11^2\\x^2=170\\x=\sqrt{170} cm\\ = 13.04cm$

c. In this question you have to find x and y. We need to find x first using pythagoras

$x^2=6^2+10^2\\x^2=136\\x=\sqrt{136}cm \\=11.66cm$

Now that we've found x we can find y using pythagoras but instead of find c, we will find another side

$c^2=a^2+b^2\\19^2=\sqrt{170} ^2 + b^2\\361=170+b^2\\361-170=170+b^2-170\\191=b^2\\b=\sqrt{191}cm \\=13.82cm$

Hope this helps :)

8 0

2 years ago