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4 years ago

Identify the binomial A. x2 −4x + 2 B. x2 C. x2 − 4 D. 2x2 + 3x −1 + y

Mathematics

2 answers:

Kamila [148]4 years ago

4 0

The correct binomial is C.

Sidana [21]4 years ago

3 0

The Correct answer is C) x² - 4 .

A binomial is an expression that contains the sum of two distinct terms. In this case, the terms are x² and -4.

A) Contains 3 terms: x², -4x, and 2. Notice that, since x and x² have different exponents, they are considered distinct terms (you can't simplify the expression any further, unless you know what the value for x is)

B) Contains only one term, x²

D) Contains 4 distinct terms, 2x², 3x, -1, and y

You might be interested in

The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width

blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given perimeter

perimeter of rectangle = 2 ( length + width )

=> 30.8 km = 2 ( x + y )

=> $x + y = \frac{30.8}{2}$

=> y = 15.4 – x ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,

=> $length^2 + width^2 = diagonal^2$

=> $x^2 + y^2 = 11^2$

=> $x^2 + y^2 = 121$

On substituting value of y from (1) in above equation we get

=> $x^2 + (15.4-x)^2 = 121$

=> $x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x = 121$

=> $2x^2-30.8x + 237.16 -121 = 0$

=> $2x^2-30.8x + 116.16 = 0$

Solving above quadratic equation using quadratic formula

General form of quadratic equation is

$ax^2 +bx +c = 0$

And quadratic formula for getting roots of quadratic equation is

$x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}$

As equation is $2x^2-30.8x + 116.16 = 0$ , in our case

a = 2 , b = -30.8 and c = 116.16

Calculating roots of the equation we get

$x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}$

$x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}$

$x=\frac{30.8\pm\sqrt{860.64}}{4}$

$x=\frac{(30.8\pm29.33)}4$

$x=\frac{(30.8+29.33)}{4}$

$x=\frac{(30.8-29.33)}{4}$

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0

3 years ago

Is this correct?- <br><br> Also please don’t just say “yes” or “no”-

hichkok12 [17]

Answer:

a) No

b) No

c) No

d) Yes

3 0

3 years ago

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

The circle graph below represents the opinions of 100 students about their favorite sports. Each student chose exactly one of th

bagirrra123 [75]

Answer:

Yo…what’s the graph?

Step-by-step explanation:

8 0

3 years ago

The Royal Fruit Company produces two types of fruit drinks. The first type is 55% pure fruit juice, and the second type is 80% p

Stolb23 [73]

Answer:

First type of fruit drinks: 48 pints

Second type of fruit drinks: 32 pints

Step-by-step explanation:

Let's call A the amount of first type of fruit drinks. 55% pure fruit juice

Let's call B the amount of second type of fruit drinks. 80% pure fruit juice

The resulting mixture should have 65% pure fruit juice and 80 pints.

Then we know that the total amount of mixture will be:

$A + B = 80$

Then the total amount of pure fruit juice in the mixture will be:

$0.55A + 0.8B = 0.65 * 80$

$0.55A + 0.8B = 52$

Then we have two equations and two unknowns so we solve the system of equations. Multiply the first equation by -0.8 and add it to the second equation:

$-0.8A -0.8B = -0.8*80$