All categories

2 years ago

ZC is an altitude, m∠CYW (5x + 22)°, and m∠WZC (12x)°. Find m∠WZC

Mathematics

1 answer:

hammer [34]2 years ago

5 0

Answer:

Step-by-step explanation:

ZC is an altitude, thus ∠ZCY = 90°

The sum of the3 angles in a triangle = 180°

Sum the 3 angles in ΔCYZ and equate to 180

90 + 5x + 22 + 12x = 180 ← simplify left side

17x + 112 = 180 ( subtract f112 from both sides )

17x = 68 ( divide both sides by 17)

x = 4

Hence

∠WZC = 12x = 12 × 4 = 48° → C

You might be interested in

Identify the values a, b, and c is the first step in using the quadratic formula to find the solution(s) to a quadratic equation

zlopas [31]

The values are a = 7, b = -9, c = -18.

<u>Step-by-step explanation:</u>

The given quadratic equation is $7x^{2} - 9x = 18$

The general form of the quadratic equation is $ax^{2} + bx + c = 0$

where,

a is the coefficient of x².
b is the coefficient of x.
c is the constant term.

Now, you have to modify the given quadratic equation similar to the general form of quadratic equation.

So, bring the constant term 18 to the left side of the equation for equating it to zero.

⇒ $7x^{2} - 9x - 18 = 0$

Compare the above equation with general form $ax^{2} + bx + c = 0$

⇒ a = 7

⇒ b = -9

⇒ c = -18

Therefore, the values of a, b, and c are 7, -9 and -18.

7 0

3 years ago

Read 2 more answers

Whats the area of the real Bedroom?

Gemiola [76]

Answer:

The answer is 18 cm.

Step-by-step explanation:

Just multiply 6 by 3.

3 0

2 years ago

Which of the following is a result of shifting a circle with equation (x + 3)2 + (y - 2)2 = 36 up 3 units? A. The x-coordinate o

Katen [24]

Answer:

D. (y-2) becomes (y-5) and -5 < -2

Step-by-step explanation:

When transforming functions, the following applies:

• Adding/subtracting inside the parenthesis to the input shifts the function left(+) and right(-).

• Adding/subtracting outside the parenthesis to the output shifts the function up(+) and down(-).

• Multiplying the function by a number less than 1 compresses it towards the x-axis.

• Multiplying the function by a number greater than 1 stretches it away from the x-axis.

In this situation, the circle is shifted up 3 units and the variable y which controls this is in the function. To move it up you will subtract 3 in the parenthesis for (y-2) so it becomes (y-5). This will move the vertex 3 units higher.

3 0

2 years ago

Read 2 more answers

Plss. pa answers ?<br><br>thanks po<br><br>

Lana71 [14]

Answer:

<h3>Graph 3</h3>

Line starting at x = -2

<u>Domain</u>: x ≥ -2
<u>Range</u>: y ≥ 0

<h3>Graph 4</h3>

Vertical line

<u>Domain</u>: x = 3
<u>Range</u>: y = any real number

<h3>Graph 5</h3>

Quadratic function with negative leading coefficient and max value of 3

<u>Domain</u>: x = any real number
<u>Range</u>: y ≤ 3

<h3>Graph 6</h3>

Curve with non-negative domain and min value of -2

<u>Domain</u>: x ≥ 0
<u>Range</u>: y ≥ -2

<h3>Graph 7</h3>

Line with no restriction

<u>Domain</u>: x = any real number
<u>Range</u>: y = any real number

<h3>Graph 8</h3>

Quadratic function with positive leading coefficient and min value of 4

<u>Domain</u>: x = any real number
<u>Range</u>: y ≥ 4

<h3>Graph 9</h3>

Parabola with restriction at x = -4

<u>Domain</u>: x = any real number except -4
<u>Range</u>: y = any real number

<h3>Graph 10</h3>

Square root function with star point (2, 0)

<u>Domain</u>: x ≥ 2
<u>Range</u>: y ≥ 0

5 0

2 years ago

HELP ME PLEASE!

shusha [124]

Answer:

Part A) The probability is closer to 0, because is less than 50%

Part B) The probability is closer to 1, because is greater than 50%

Step-by-step explanation:

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.

The size of the sample space is the total number of possible outcomes

The event space is the number of outcomes in the event you are interested in.

Let

x------> size of the event space

y-----> size of the sample space

$P=\frac{x}{y}$

Part A) Is the probability of hitting the black circle inside the target closer to 0 or 1?

In this problem we have

x------> is the area of the black circle

y------> is the total area of the square target

<em>Find the area of black circle</em>

$r=3/2=1.5\ units$

$x=(3.14)(1.5)^{2} =7.065\ units^{2}$

<em>Find the total area of the square target</em>

$y=9^{2}=81\ units^{2}$

Find the probability

$P=\frac{7.065}{81}=0.0872$

Convert to percentage

$0.0872*100=8.72\%$

The probability is closer to 0, because is less than 50%

Part B) Is the probability of hitting the white portion of the target closer to 0 or 1?

In this problem we have

x------> is the area of the white portion of the target

y------> is the total area of the square target

<em>Find the area of the white portion</em>

The area is equal to the area of the square minus the area of the circle

$x=81-7.065=73.935\ units^{2}$

Find the probability

$P=\frac{73.935}{81}=0.9128$

Convert to percentage

$0.9128*100=91.28\%$

The probability is closer to 1, because is greater than 50%

4 0

2 years ago