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1 year ago

Please help i dont get this at all so please i will highly appreciate it .Thank youuu!!!

Mathematics

1 answer:

svlad2 [7]1 year ago

6 0

Answer:

Step-by-step explanation:

definitely

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4. What are the terms in the expression, 9- 4 + 3b + 7a?

Elan Coil [88]

Answer:

The terms are 9, 4, 3b and 7a.

The terms are 9 – 4 and 3b + 7a.

The terms are 9, -4, 3b, and 7a.

Step-by-step explanation:

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2 years ago

What Is The scale factor of the dilation? A)-1/2 B)1/2 C)-2 D)2

Vadim26 [7]

Answer:

B) 1/2

Step-by-step explanation:

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3 years ago

The function p(x) = −2000x2 + 18000x − 15000 models the monthly profit P of a smoothie company, where x represents the price per

ohaa [14]

Answer:

Step-by-step explanation:

Givem the profit function

p(x) = −2000x2 + 18000x − 15000

We are to generate the price range that will generate a monthly profit of at least $25,000

Substitute into the function we have;

25000 = −2000x2 + 18000x − 15000

Divide through by 1000

25 = -2x²+18x-15

Rearrange

-2x²+18x-15-25 = 0

2x²-18x+40 = 0

Divide through by 2

x²-9x+20 = 0

Factorize

x²-5x-4x+20 = 0

x(x-5)-4(x-5) = 0

x-4 = 0 and x-5 = 0

x = 4 and x = 5

Hence the price range that will generate a monthly profit of at least $25,000 is between $4 and $5 inclusive

8 0

2 years ago

Please help!! I don't know what to do :(

elena55 [62]

Yo sup??

This question can be solved by applying the properties of similar triangles

the triangle with sides 5x and 20 is similar to the triangle with sides 45 and 36

therefore we can say

5x/45=20/36

x=5 units

Hope this helps

8 0

2 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).

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