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2 years ago

Which statement is not a step used when constructing an inscribed square?

1 answer:

3 0

The statement that is not a step used when constructing an inscribed square is swing an arc the length of the radius from the point on the circle. Option B

<h3>Steps in inscribing a square</h3>

The steps involved in inscribing a square in a square are;

Use a compass to draw a circle and label the center
Draw a diameter of the circle using a straightedge
Then, construct the perpendicular bisector of the diameter
Label the points where the bisector intersects the circle
Connect the points to form the square.

From the above steps, it can be concluded that swinging an arc the length of the radius from the point on the circle is not a step in inscribing a square.

Hence, the statement that is not a step used when constructing an inscribed square is swing an arc the length of the radius from the point on the circle. Option B

Learn more about an inscribed square here:

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Solve for x.<br><br> x - 4.225 = 15.1

MArishka [77]

Answer:

x = 19.325

Step-by-step explanation:

You have: x - 4.225 = 15.1

Isolate x by adding 4.225 to both sides

x - 4.225 + 4.225 = 15.1 + 4.225

x = 15.1 + 4.225 = 19.325

x = 19.325

8 0

3 years ago

A factory received a shipment of 11 hammers, and the vendor who sold the items knows there are 2 hammers in the shipment that ar

zhenek [66]

Question:

If a sample of 2 hammer is selected

(a) find the probability that all in the sample are defective.

(b) find the probability that none in the sample are defective.

Answer:

a $Pr = \frac{2}{110}$

b $Pr = \frac{72}{110}$

Step-by-step explanation:

Given

$n = 11$ --- hammers

$r = 2$ --- selection

This will be treated as selection without replacement. So, 1 will be subtracted from subsequent probabilities

Solving (a): Probability that both selection are defective.

For two selections, the probability that all are defective is:

$Pr = P(D) * P(D)$

$Pr = \frac{2}{11} * \frac{2-1}{11-1}$

$Pr = \frac{2}{11} * \frac{1}{10}$

$Pr = \frac{2}{110}$

Solving (b): Probability that none are defective.

The probability that a selection is not defective is:

$P(D') = \frac{9}{11}$

For two selections, the probability that all are not defective is:

$Pr = P(D') * P(D')$

$Pr = \frac{9}{11} * \frac{9-1}{11-1}$

$Pr = \frac{9}{11} * \frac{8}{10}$

$Pr = \frac{72}{110}$

8 0

3 years ago

without building the graph, find the coordinates of the point of intersection of the lines given by the equation y=3x-1 and 3x+y

DaniilM [7]

<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

$y = 3x - 1$
$3x + y = -7$

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

$\implies 3x + y = -7$

$\implies3x + (3x - 1) = -7$

$\implies3x + 3x - 1 = -7$

Combine like terms as needed;

$\implies 3x + 3x - 1 = -7$

$\implies 6x - 1 = -7$

Add 1 to both sides of the equation;

$\implies 6x - 1 + 1 = -7 + 1$

$\implies 6x = -6$

Divide 6 to both sides of the equation;

$\implies \dfrac{6x}{6} = \dfrac{-6}{6}$

$\implies x = -1$

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

$\implies y = 3(-1) - 1$

$\implies$ $\ \ = -3 - 1$

$\implies$ $= -4$

Therefore, the value(s) of the x-variable and the y-variable are;

$\boxed{x = -1}$ $\boxed{y = -4}$

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

$\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.$

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

$\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

$\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

$\implies 0 = (6x) + (6)$

$\implies0 = 6x + 6$

This problem is now an algebraic problem. Isolate "x" to determine its value.

$\implies 0 - 6 = 6x + 6 - 6$

$\implies -6 = 6x$

$\implies -1 = x$

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

$\implies y = 3(-1) - 1$

$\implies y = -3 - 1$

$\implies y = -4$

Therefore, the value(s) of the x-variable and the y-variable are;

$\boxed{x = -1}$ $\boxed{y = -4}$

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

5 0

2 years ago

Are the two triangles similar? how do you know

Zinaida [17]

I forgot what they’re called but the angles opposite to each other are the same degree. It’s a theorem, just search up the name

7 0

3 years ago

A landscape contractor planted 37 tulip bulbs, 53 daffodil bulbs, and 82 crocus bulbs in each field. He planted bulbs in 174 fie

ohaa [14]

He planted D. 29,928 Bulbs
First Add up all the bulbs.
Then multiply the # of bulbs with the # of fields.

3 0

3 years ago