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3 years ago

15

I need the help as possible please somebody help me I need help I swear please somebody help me help me help me please someone h

elp me please I swear I need the help from somebody please

1 answer:

muminat3 years ago

3 0

Here is your answer

A. (4,-7)

REASON:

In quadrant IV abscissa is +ve and ordinate is -ve. Since absolute value is 11

Here |4|+|-7|= 4+7= 11

HOPE IT IS USEFUL

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3 years ago

Tara has 4 4/5 feet of rope. Part A: How many 2/5 foot pieces can Tara cut from the 4 4/5 feet of rope? Show your work. Part B:

aleksley [76]

Answer:

Part A

12 pieces

Part B

Tara can cut Twelve 2/5 foot pieces can be cut from 4 4/5 feet of rope.

Step-by-step explanation:

Part A: How many 2/5 foot pieces can Tara cut from the 4 4/5 feet of rope?

This is calculated as:

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= 24/5 × 5/2

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Part B: Using the information in Part A, interpret the meaning of the quotient in terms of the two fractions given

The quotient in Part A is 12

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3 years ago

The sum of four consecutive integers are 198. What's the fourth number in this sequence?

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3 years ago

How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>2</u>

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u> $\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$ </u>

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u> $\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$ </u>

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>4</u>

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>3</u>

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>4</u>

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

2 years ago