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3 years ago

9

Hey can i get some help on this question?

2 answers:

vredina [299]3 years ago

7 0

which are the optiones?

Goryan [66]3 years ago

5 0

5 times:

for example, when X=1, y=5, and when you look at the relationship between 1 and 5, 1 times 5 is 5. So, y is 5 times of x.

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Sandra went for pizza with friends. She ordered a slice of pizza with a side salad for $9.50. The tax was 6.2%.

Anvisha [2.4K]

The total amount of money Sandra paid for a slice of pizza with a side salad including tax is $10.089

<h3>How to find total price including tax</h3>

Price of slice of pizza with a side salad = $9.50
Tax = 6.2%

Total amount Sandra paid = Price of slice of pizza with a side salad + (Tax × Price of slice of pizza with a side salad)

= 9.50 + (6.2% × 9.50)

= 9.50 + (0.062 × 9.50)

= 9.50 + 0.589

= $10.089

Therefore, the total amount Sandra paid is $10.089

Learn more about percentage:

brainly.com/question/843074

3 0

2 years ago

The greatest common factor of 2 different prime numbers is 1

Georgia [21]

Answer:

ok is this supposed to be a quesion to ask or a fact?

Step-by-step explanation:

4 0

3 years ago

How do you solve 6650÷50

SpyIntel [72]

Answer:

its 133

Step-by-step explanation:

50 fits into 6650 133 times. u cant show a pic but if you know division, its easy.

4 0

3 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

3 years ago

A circle with radius 4 inches is inscribed in an equilateral triangle. Find the area of the triangle.

LUCKY_DIMON [66]

The inscribed circle has its center at the point of intersection of the angle bisectors, which also happen to be the medians. Hence the altitude of the triangle is 3 times the radius, or 12 inches.

The side length of this triangle is 2/√3 times the altitude, so the area is

... Area = (1/2)·b·h = (1/2)·(24/√3 in)·(12 in)

... Area = 48√3 in² ≈ 83.1384 in²

7 0

3 years ago