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2 years ago

Help find the value of x

Mathematics

1 answer:

Masteriza [31]2 years ago

8 0

Answer: i believe

that x does not have a value

Step-by-step explanation:"variable" or sometimes an "unknown"

You might be interested in

F(x) = 49 - x^2 (squared)<br><br> f(5) = ___

Ostrovityanka [42]

Answer: 24

Step-by-step explanation:

we replace x with 5

f(5)=49-25=24

3 0

2 years ago

Determine two pairs of polar coordinates for the point (4, -4) with 0° ≤ θ < 360°.

Ray Of Light [21]

Answer:

Step-by-step explanation:

$4=r cos \theta\\-4=r sin \theta\\square ~and~add\\16+16=r^2(cos^2 \theta+sin^2\theta)\\r^2=32\\ r=4\sqrt{2} \\divide \\tan \theta=-1\\as x is positive ,y is negative ,so \theta lies in 4th quadrant.\\tan \theta=-1=-tan 45=tan(360-45)=tan 315\\\theta=315°\\\\co-ordinates~ are~(r,theta) ~or~(-r,\theta+ -180°)\\hence ~co-ordinates~are(4\sqrt{2} ,315°),(-4\sqrt{2} ,135°)$

7 0

3 years ago

: Find the area of the shaded region for these two questions. Remember this

xxMikexx [17]

Answer:

c) 18 cm²

d) 30 cm²

Step-by-step explanation:

You can do this 2 different ways. Either way, you need to use the bottom side of 8 cm and the two congruent segments. Each of the two congruent segments of the bottom side measures 4 cm.

Method 1) Find the area of the large triangle and subtract from it the area of the small triangle.

area of triangle = bh/2

shaded area = 8 cm × 6 cm / 2 - 4 cm × 3 cm / 2

shaded area = 24 cm² - 6 cm²

shaded area = 18 cm²

Method 2) The shaded region is a trapezoid with parallel bases of lengths 3 cm and 6 cm and height 4 cm. Find the area of the trapezoid.

area of trapezoid = (B + b)h/2

shaded area = (6 cm + 3 cm)(4 cm)/2

shaded area = (9 cm)(4 cm)/2

shaded area = 18 cm²

The area of the shaded region is the area of the white triangle subtracted from the area of the rectangle.

area of rectangle = length × width

area of triangle = bh/2

shaded area = (12 cm)(4 cm) - (9 cm)(4 cm)/2

shaded area = 48 cm² - 18 cm²

shaded area = 30 cm²

6 0

2 years ago

PLEASE HELP ASAP: TRIGONOMETRY

ratelena [41]

Recall that sine is the ratio of opposite over hypotenuse, so we say

sin(theta) = opposite/hypotenuse

If we have a triangle labeled as such in the diagram below, then we can use symbols to write sin(theta) = A/C

Similarly, cos(theta) = B/C since cosine is the ratio of adjacent over hypotenuse.

Then from here we use a bit of algebra to compute the left hand side of the given equation. The goal is to see if we can simplify it to 1.

$\sin^2\left(\theta\right)+\cos^2\left(\theta\right)\\\\\left(\sin\left(\theta\right)\right)^2+\left(\cos\left(\theta\right)\right)^2\\\\\left(\frac{A}{C}\right)^2+\left(\frac{B}{C}\right)^2\\\\\frac{A^2}{C^2}+\frac{B^2}{C^2}\\\\\frac{A^2+B^2}{C^2}\\\\\frac{C^2}{C^2}\\\\1$

So this shows that the left hand side is identically the same as the right hand side (both sides are 1), and it doesn't matter what you replace theta with. Theta can be any real number and the equation will always be true.

Note: if you're wondering how A^2+B^2 became C^2, it's because of the pythagorean theorem. So that's why this trig identity is often referred to as the pythagorean trig identity.

6 0

3 years ago

A number is selected from the set {1, 2, 3, 5, 15, 21, 29, 38, 500}. If equal elemental probabilities are assigned, what is the

Zanzabum

Answer:

The required probability is $\frac{7}{9}$ .

Step-by-step explanation:

Set given is:

S = {1, 2, 3, 5, 15, 21, 29, 38, 500}

Total number of elements in set, $n(S)$ = 9

Let A be the event that the number is less than 29 ({1, 2, 3, 5, 15, 21}).

Number of items in the event A, $n(A)$ = 6

Probability of event A,

$P(A) = \dfrac{n(A)}{n(S)}}=\dfrac{6}{9} \Rightarrow \dfrac{2}{3}$

Formula for probability of any event E:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

Let B be the event that the number is odd (either of {1,3,5,15,21,29}).

Number of items in the event B, $n(B)$ = 6

Probability of event B,

$P(B) = \dfrac{n(B)}{n(S)}}=\dfrac{6}{9} \Rightarrow \dfrac{2}{3}$

The event A and B have a few elements in common, i.e. numbers less than 29 which are odd as well.

The common elements are represented as:

$A \cap B = \{1, 3, 5, 15, 21\}$

$n(A\cap B) = 5$

$P(A \cap B ) = \dfrac{n(A \cap B)}{n(s)}\\\Rightarrow P(A \cap B) = \dfrac{5}{9}$

To find probability of selecting a number which is either less than 29 (event A) or odd (event B),

We have to find $P(A\ or \ B)$ which is represented as $P(A \cup B)$ and the formula is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)\\\Rightarrow \dfrac{2}{3} + \dfrac{2}{3} - \dfrac{5}{9}\\\Rightarrow \dfrac{12-5}{9}\\\Rightarrow \dfrac{7}{9}$

The required probability is $\frac{7}{9}$ .

7 0

3 years ago