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Kitty [74]
3 years ago
10

Gerardo says that a cube with edges that measure 10 centimeters has a volume that is twice as much as a cube with sides that mea

sure 5 centimeters. Explain and correct Gerardo’s error.
Mathematics
1 answer:
irga5000 [103]3 years ago
5 0
He is wrong.
The 10 inch cube is bigger than the 5 inch cube by 5 cm in EACH DIRECTION.
So, the 10 inch cube should be twice as big in volume by a factor of 2*2*2
=8
Okay let's see.
5 cm cube volume = 5*5*5 = 125 cc
10 cm cube volume = 10 * 10 * 10 = 1,000 cc

1,000 / 125 = 8



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Find 0.<br>Trigonometry problems.
aivan3 [116]

In the triangle with legs 7 and 9 cm, you can find the length of the hypotenuse by using the Pythagorean theorem. If the hypotenuse is x, then

7^2+9^2=x^2\implies x=\sqrt{130}

The other right triangle with leg 6 cm and containing the angle \theta satisfies

\sin\theta=\dfrac{6\,\rm cm}{\sqrt{130}\,\rm cm}\implies\theta=\sin^{-1}\dfrac6{\sqrt{130}}\approx31.75^\circ

6 0
3 years ago
Error Analysis! Answer both parts of the question.
Maurinko [17]

Answer:

Kaitlin's answer is wrong because she is supposed to by dividing. When trying to get rid of a number from a variable you have to divide the number by itself and whatever you do to one side you have to do to the other side

so you would do 5/5 - cancels out

15/5=3

y=3

Step-by-step explanation:

have a great day :D

8 0
2 years ago
I need it done asapp<br> Will give brainly!!
FinnZ [79.3K]

Answer:

I believe the answer is 1/5 please tell me if I am wrong and I am truly sorry if I am

Step-by-step explanation:

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%
andrey2020 [161]

First observe that if a+b>0,

(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}

Let a=0 and b=x. It follows that

a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}

Now let b=1, so a^2+a=4x. Solving for a,

a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2

which means

a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}

Now solve for x.

x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2

(note that we assume 2x-1\ge0)

4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}

(we omit x=0 since 2\cdot0-1=-1\ge0 is not true)

3 0
1 year ago
5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is
Dvinal [7]

Answer:

  x = -2 or -9

Step-by-step explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

<h3>Slope</h3>

The slope of a line is given by the slope formula:

  m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

  m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

  m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

<h3>Perpendicular lines</h3>

The slopes of perpendicular lines have product of -1:

  \dfrac{x}{2x+3}\cdot\dfrac{2+3x}{6-x}=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}

<h3>Solutions</h3>

The values of x that satisfy this equation are x = -2 and x = -9. The attached graphs show the lines for each of these cases.

4 0
1 year ago
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