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3 years ago

13

I need help and can u please show your work please and thank u

1 answer:

aivan3 [116]3 years ago

6 0

Answer:

Ur mom lol she gave birth to a kid who knows nothin

Step-by-step explanation:

You might be interested in

28 in 21 in., 20 in.<br> Is this a right triangle?

ankoles [38]

Answer:

No, it is not a right triangle.

Step-by-step explanation:

The simplest way to determine is testing out the numbers with Pythagorian theorem.

If it complies with the theorem, it is a right triangle.

let's assume c = 28, b = 21, and a = 20

the longest side is the hypotenuse so side c (28 in) will be the hypotenuse.

According to the Pythagorian theorem, the square of the length of hypotenuse must equal to the sum of squares of other two sides.

check:

c^2 = 28^2 = 784

a^2 + b^2 = 21^2 + 20^2 = 841

because c^2 is not equal to a^2 + b^2, the triangle is not a right triangle.

7 0

2 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

2 years ago

Use the square root property of equality to solve (x – 3)2 = –4. The solutions are a. -1 or 7 b. 1 or 5 c. 3+- 2i d. 3+-4i

makkiz [27]

The answer is C. 3+-2i

(x-3) $(x-3)^{2} = -4

 \sqrt{(x-3)^2} = +- \sqrt{-4

x - 3 = i \sqrt{4} 

x - 3 = +-2i

x = 3 +-2i$

6 0

2 years ago

Read 2 more answers

On Monday, a local hamburger shop sold a combined total 288 of hamburgers and cheeseburgers. The number of cheeseburgers sold wa

Oxana [17]

Answer:

288=x+(2x)

288=96+(2x96)

288=96+(192)

288=288

x=96

can I have a brainliest lol

3 0

2 years ago

All cars have a speedometer and an odometer. Which one measures a rate of change? What are the two variables in this rate of cha

ryzh [129]

Answer:

Step-by-step explanation:

The speedometer measures the rate of change of distance with respect to time.

The odometer displays the total distance traveled since it was set to 0.00.

3 0

3 years ago