All categories

2 years ago

Help plz! I suck at math, and I have already failed this quiz 2 times.

Mathematics

1 answer:

kondaur [170]2 years ago

8 0

It’s A yw loves....

You might be interested in

If y = 4x + 0.5, find y when x =1.5

inn [45]

Answer:

= 6.5

Step-by-step explanation:

You have to plug in 1.5 for x

4(1.5) + 0.5 = 6.5

8 0

2 years ago

How to solve 2x(x+3)=2x^2+15

Travka [436]

Answer: x = 5/2 = 2.5

Step-by-step explanation:

<u>Given expression</u>

2x (x + 3) = 2x² + 15

<u>Expand parentheses and apply the distributive property</u>

2x · x + 2x · 3 = 2x² + 15

2x² + 6x = 2x² + 15

<u>Subtract 2x² on both sides</u>

2x² + 6x - 2x² = 2x² + 15 - 2x²

6x = 15

<u>Divide 6 on both sides</u>

6x / 6 = 15 / 6

x = 15/6

$\boxed{x=\frac{5}{2} }$

Hope this helps!! :)

Please let me know if you have any questions

5 0

2 years ago

Read 2 more answers

X^2+y^2+14x+10y=7 <br> center and radius

Sholpan [36]

$(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-the\ coordinates\ of\ the\ center\\r-the\ radius\\\\==================================\\Use:(k+l)^2=k^2+2kl+l^2\ (*)\\==================================$

$x^2+y^2+14x+10y=7\\\\x^2+14x+y^2+10y=7\\\\x^2+2x\cdot7+y^2+2y\cdot5=7\\\\\underbrace{x^2+2x\cdot7+7^2}_{Use\ (*)}-7^2+\underbrace{y^2+2y\cdot5+5^2}_{Use\ (*)}-5^2=7\\\\(x+7)^2-49+(y+5)^2-25=7\\\\(x+7)^2+(y+5)^2-74=7\ \ \ \ |add\ 74\ to\both\ sides\\\\(x+7)^2+(y+5)^2=81\\\\(x+7)^2+(y+5)^2=9^2$

$Answer:\\\boxed{(-7;-5)-center;\ 9-radius}$

8 0

3 years ago

Read 2 more answers

A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is ce

lubasha [3.4K]

<h2>Answer:</h2>

$\boxed{(5(-\sqrt{2}+\sqrt{6}),5(\sqrt{2}+\sqrt{6}))}$

<h2>Step-by-step explanation:</h2>

The origin of the coordinate system is the center of the circle. So we have an angle that measures $5\pi/12=75^{\circ}$ . so the x-coordinate and y-coordinate can be found, by using trigonometry as follows:

$x=20cos(5\pi/12)=5\sqrt{6}-5\sqrt{2} \\ \\ Arranging:\\ \\x=5(-\sqrt{2}+\sqrt{6}) \\ \\y=20sin(5\pi/12)=5\sqrt(6)+5\sqrt{2} \\ \\ Arranging:\\ \\y=5(\sqrt{2}+\sqrt{6})$

Finally, the exact value of the position of the rider after the carousel rotates $5\pi/12$ radians is:

$\boxed{(5(-\sqrt{2}+\sqrt{6}),5(\sqrt{2}+\sqrt{6}))}$

3 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