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3 years ago

Choose the simplified solution(s) to this equation: ?- 128 - 0 18 1642 V128 18/2 Question 33 )

Mathematics

1 answer:

tatiyna3 years ago

7 0

D is correct answer

±8√2

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Leftover Expression

Mumz [18]

The second leftover expression is not o(a+b). It is 6(a + b). I have attached the correct question to depict that.

Answer:

The equivalent expressions are;

8a + 2 and 6a + 6b

Step-by-step explanation:

The two leftover expressions are given as;

2(4x + 1) and 6(a + b)

In algebra, equivalent expressions are simply those expressions which when simplified, give the same resulting expression as the initial one.

Thus simply means expanding or collecting like times to make it clearer.

Now, in our question, like terms have already been collected. This means that to find an equivalent expression, we will just expand the bracket.

Thus;

2(4x + 1) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;

8x + 2

Similarly,

6(a + b) will be expanded by using the 2 outside the bracket to multiply the terms inside the bracket. This gives;

6a + 6b

Thus;

The equivalent expressions are;

8a + 2 and 6a + 6b

4 0

3 years ago

MATH QUESTION SCREEN SHOT DOWN BELOW

Lisa [10]

F(x)= (x-p)²+q where (p,q) is min point
f(x)=(x-1)²-2
min point is (1,-2)
the answer is first picture

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

3 years ago

Rick wants to read 52 books his year.he has already read 24 books.how many more should he read?

stira [4]

The answer is 28 and you are welcome

7 0

3 years ago

Read 2 more answers

Use the figure below to answer the question Solve for x

musickatia [10]

Answer:

Step-by-step explanation:

The sum of the interior angles, in degrees, of a regular polygon is given by the formula 180(n – 2), where n is the number of sides. The problem concerns a polygon with six sides, so we will let n = 6. The sum of the interior angles in this polygon would be 180(6 – 2) = 180(4) = 720. The angle on the bottom left is a right angle because of the square indicator. We know that this is 90 degrees. If we take the sum of the angles that we know we get 593. 720 - 593 = 127. x = 127.

6 0

2 years ago

Read 2 more answers

Choose the simplified solution(s) to this equation: ?- 128 - 0 18 1642 V128 18/2 Question 33 )​

Choose the simplified solution(s) to this equation: ?- 128 - 0 18 1642 V128 18/2 Question 33 )