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2 years ago

Please help. I really don't seem to understand this question.

Mathematics

1 answer:

ivann1987 [24]2 years ago

3 0

Answer:

m3=15(8)+22=142

m1=19(8)-10=142

Both are equal so the answer is Yes they are parallel

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Mila jogged 4.6 miles in 48 minutes. if she jogged at a constant rate what was her jogging speed.

butalik [34]

Answer: 5.75 miles per hour

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Need help with this question.

Vaselesa [24]

Answer:

4:1 (I think)

Step-by-step explanation:

Let's write some equations first:

For both prices increasing by 20:

$\frac{x+20}{y+20} =\frac{5}{2}$

then we can use cross multiplication

2x+40=5y+100

For both prices decreasing by 5:

$\frac{x-5}{y-5} =\frac{5}{1}$

same here:

5y-25=x-5

Using substitution for 5y, we get: 5y=x+20

2x+40=5y+100

2x+40=x+20+100

x=80

plugging this in to the second equation:

5y-25=x-5

5y-25=80-5

5y-25=75

5y=100

y=20

therefore we get x:y=80:20

4:1

4 0

2 years ago

The graph of f(x) = x3 − 7x − 6 is shown. Based on the graph, what are all of the solutions to f(x) = x3 − 7x − 6? x = −6 x = −2

pogonyaev

F(x)=x^3-7x-6 Since I don't have the graph and this is not a perfect cube, I will have to rely on Newton :P

x-(f(x)/(dy/dx))

x-(x^3-7x-6)/(3x^2-7)

(2x^3+6)/(3x^2-7), letting x1=0

0, -6/7, -.988, -.9999, -.99999999999, -1

(x^3-7x-6)/(x+1)

x^2 r -x^2-7x-6
-x r -6x-6
-6 r 0

(x+1)(x^2-x-6)=0

(x+1)(x-3)(x+2)=0

x= -2, -1, 3

4 0

3 years ago

Read 2 more answers

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).

First Question

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.

Find Q2

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

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

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

Second Question

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.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

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

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:-

For Q3

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

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

For Q4

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

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

What is 154 plus 74 equal

Sergio [31]

154 + 74 = 228 is the answer.

3 0

3 years ago

Read 2 more answers

Please help. I really don't seem to understand this question. ​

Please help. I really don't seem to understand this question.