If a polynomial function f(x) has roots 3 and square root of 7, what must also be a root of f(x)

C

By the law of sines, [tex]\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}[/tex] where A, B, C are the angles and a, b, c are the lengths of the sides opposite their respective angles. In this case, [tex]110^{\circ}[/tex] is opposite 4.6 and A is opposite 3.2, so [tex]\frac{sinA}{3.2}=\frac{sin(110^{\circ})}{4.6}[/tex], giving the answer.

Answer:

it’s c

Step-by-step explanation:

Answer:

−5 + 5 = 0

-6+6=0

14+-14=0

70+-70=0

100+-100=0

Step-by-step explanation:

hope this helps

Answer:

8/17

Step-by-step explanation:

Since this is a right triangle, we can use the trigonometric identities

sin Y = opposite side/ hypotenuse

         = 16/34

Dividing the top and bottom by 2

          = 8/17

Answer:

B

Step-by-step explanation:

A secant is a line which intersects a circle at 2 points

From the diagram this is XY → B

The correct option is option B: The line segment XY is a secant.

Line segment which intersects the circle at two points is called secant.

We are given circle with center at O.

From the options,

1.  Line UZ :

   UZ intersects the circle at one point. So it can't be secant.

2. Line XY:

    XY intersects the circle at TWO points. So it is a secant.

3. Line XO :

   XO intersects the circle at one point. So it can't be secant.

Therefore the line segment XY is a secant.

Learn more about secant

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Answer:

a=3 b=-2 c=4

Step-by-step explanation:

Answer:

Step-by-step explanation:

A function is a process or a relation that associates each element x of a set X, to a single element y of another set Y.

We have:

{(3, -2), (1, 2), (-1, -4), (-1, 2)}

for x = -1 are two values of y = -4 and y = 2. Therefore this realtion is not a function.

Answer: No, it is not a function.

Step-by-step explanation:

The given relation:  {(3, −2), (1, 2), (−1, −4), (−1, 2)}

According to the above definition, the given relation is not a function because -1 corresponds to two different output values i.e. -4 and 2.

Hence, the given relation is not a function.

Answer:

The graph in the attached figure ( is the third option)

Step-by-step explanation:

we have the compound inequality

[tex]-18> -5x+2\geq -48[/tex]

Divide the compound inequality in two inequalities

[tex]-18> -5x+2[/tex] -----> inequality A

[tex]-5x+2\geq -48[/tex] -----> inequality B

Step 1

Solve inequality A

[tex]-18> -5x+2[/tex]

[tex]-18-2> -5x[/tex]

[tex]-20> -5x[/tex]

Multiply by -1 both sides

[tex]20<5x[/tex]

[tex]4<x[/tex]

Rewrite

[tex]x > 4[/tex]

The solution of the inequality A is the interval ------>(4,∞)

Step 2

Solve the inequality B

[tex]-5x+2\geq -48[/tex]

[tex]-5x\geq -48-2[/tex]

[tex]-5x\geq -50[/tex]

Multiply by -1 both sides

[tex]5x\leq 50[/tex]

[tex]x\leq 10[/tex]

The solution of the inequality B is the interval -----> (-∞,10]

therefore

The solution of the compound inequality is

(4,∞) ∩ (-∞,10]=(4,10]

All real numbers greater than 4 (open circle) an less than or equal to 10 (close circle)

The solution in the attached figure

Answer:

a = 20 cm

Step-by-step explanation:

Since the triangle is right use Pythagoras' identity to solve for a

The square on the hypotenuse is equal to the sum of the squares on the other two sides, that is

a² + 21² = 29²

a² + 441 = 841 ( subtract 441 from both sides )

a² = 400 ( take the square root of both sides )

a = [tex]\sqrt{400}[/tex] = 20

Answer:

(x-4)(x-6)(x+3) or in more compressed form x³-7x²-6x+72

Step-by-step explanation:

To find the L.C.M, w first factorize each of the expressions.

x²-10x+24

Two numbers that when added give -10 but when multiplied give 24

will be, -4 and -6

Thus the expression becomes:

x²-4x-6x+24

x(x-4)-6(x-4)

=(x-4)(x-6)

Let us factorize the second expression.

x²-x-12

Two numbers when added give -1 and when multiplied give -12

are 3 and -4

Thus the expression becomes: x²-4x+3x-12

x(x-4)+3(x-4)

(x-4)(x+3)

Therefore the LCM between  (x-4)(x-6) and (x-4)(x+3)

will be

(x-4)(x-6)(x+3)

We can multiply the expression as follows.

(x-4)(x-6)

x²-6x-4x+24 = x²-10x+24

(x+3)(x²-10x+24)

=x³-10x²+24x+3x²-30x+72

=x³-7x²+-6x+72

Answer:

x = 142

Step-by-step explanation:

We are given the following expression for which we are to find the solution:

[tex] \sqrt { x + 2 } - 1 5 = - 3 [/tex]

Rearranging the equation to get:

[tex] \ sqrt { x + 2 } = - 3 + 1 5 [/tex]

Taking square root on both sides of the equation to get:

[tex](\sqrt{x+2} )^2=(12)^2[/tex]

[tex]x+2=144[/tex]

x = 142

[tex]\displaystyle\\\sqrt{x+2}-15=-3\\\\\sqrt{x+2}=-3+15\\\\\sqrt{x+2}=12~~\Big|~^2\\\\x+2=144\\\\x=144-2\\\\\boxed{x=142}[/tex]

.

Answer:

(E) 1

Step-by-step explanation:

2r+ (2r+3 ) < 11

Combine like terms

4r+3 < 11

Subtract 3 from each side

4r +3-3 < 11-3

4r < 8

Divide each side by 4

4r/4 < 8/4

r <2

The only possible choice is 1

Answer:

(E) 1

Step-by-step explanation:

the sum of 2r and 2r+3  less than 11 means :

2r+2r+3<11

we simplify we get :

4r+3<11

we take the 3 to the left :

4r<11-3

means

4r<8

we divide both sides by 4 we get :

r<2

so among all those values only the 1 satisfies the condition 1<2

so the answer is E

Answer:

Point A' will be located 10 units away from point A along ray PA

Step-by-step explanation:

we have

The scale factor is 3

step 1

Find out the distance PA'

we know that

The distance PA' is equal to multiply the distance PA by the scale factor

so

[tex]PA'=PA*3[/tex]

we have

[tex]PA=5\ units[/tex]

substitute the given values

[tex]PA'=(5)*3=15\ units[/tex]

step 2

Find out how many units away from point A along ray PA will A’ be located

we know that

[tex]PA'=PA+AA'[/tex]

we have

[tex]PA=5\ units[/tex]

[tex]PA'=15\ units[/tex]

substitute the given values and solve for AA'

[tex]15=5+AA'[/tex]

[tex]AA'=15-5=10\ units[/tex]

therefore

Point A' will be located 10 units away from point A along ray PA

Answer:

10 units

Step-by-step explanation:

Answer:

[tex]f(x)=10(5^{2})^{\frac{x}{2}}[/tex]  (first option)

Step-by-step explanation:

we have

[tex]f(x)=10(5)^{x}[/tex]

where

x ----> is the time in years

we know that

Crista wants to manipulate the formula to an equivalent form that calculates every half-year

The exponent will be

x/2 -----> the time every half year

To find an equivalent form

[tex]f(x)=10(5^{a})^{\frac{x}{2}}[/tex]

[tex]10(5)^{x}=10(5^{a})^{\frac{x}{2}}[/tex]

[tex]10(5)^{x}=10(5)^{a\frac{x}{2}}[/tex]

so

[tex]x={a\frac{x}{2}}[/tex]

[tex]a=2[/tex]

The equivalent form is

[tex]f(x)=10(5^{2})^{\frac{x}{2}}[/tex]

Answer:

Part A)

x=-3i

x=3i

Part B)

(x+3i)(x-3i)

Step-by-step explanation:

Given:

Part A)

x^2+9=0

x^2=-9

x= √-9

x=√-1 *√9

x=± i *3

x=±3i

Part B)

x^2+9=0

x^2 - (-9)=0

x2-(3i)^2=0

(x-3i)(x+3i)=0 !

[tex]\huge{\boxed{y-3=m(x-2)}}[/tex]

Point-slope form is [tex]y-y_1=m(x-x_1)[/tex], where [tex]m[/tex] is the slope and [tex](x_1, y_1)[/tex] is a point on the line.

Substitute the values. [tex]\boxed{y-3=m(x-2)}[/tex]

Note: This is in point-slope form. Let me know if you need a different form. Also, if you have any more problems similar to this, I encourage you to try them on your own, and ask on here if you are having trouble.

Answer:

y-3 = 3(x-2)  point slope form

y = 3x-3  slope intercept form

Step-by-step explanation:

We can use the point slope form of the equation for a line

y-y1 = m(x-x1)

where m is the slope and (x1,y1) is the point

y-3 = 3(x-2)  point slope form

If we want the line in slope intercept form

Distribute

y-3 = 3x-6

Add 3 to each side

y-3+3 = 3x-6+3

y = 3x-3  slope intercept form

Answer:

(k ∘ p)(x)=2x^2-12x+13

Step-by-step explanation:

(k ∘ p)(x)=k(p(x))

(k ∘ p)(x)=k(x-3)

(k ∘ p)(x)=2(x-3)^2-5

(k ∘ p)(x)=2(x-3)(x-3)-5

Use foil on (x-3)(x-3) or use this as a formula:

(x+a)^2=x^2+2ax+a^2.

(k ∘ p)(x)=k(p(x))

(k ∘ p)(x)=k(x-3)

(k ∘ p)(x)=2(x-3)^2-5

(k ∘ p)(x)=2(x-3)(x-3)-5

(k ∘ p)(x)=2(x^2-6x+9)-5

Distribute: multiply terms inside ( ) by 2:

(k ∘ p)(x)=2x^2-12x+18-5

(k ∘ p)(x)=2x^2-12x+13

The composite function is [tex]k(p(x))=2x^{2} -12x+13[/tex]

Option b is correct.

Given function are,

               [tex]k(x)=2x^{2} -5,p(x)=(x-3)[/tex]

We have to find composite function [tex]k(p(x))[/tex].

                  [tex]k(p(x))=k(x-3)\\\\k(p(x))=2(x-3)^{2}-5\\ \\k(p(x))=2(x^{2} +9-6x)-5\\\\k(p(x))=2x^{2} +18-12x-5\\\\k(p(x))=2x^{2} -12x+13[/tex]

Thus, the composite function is [tex]k(p(x))=2x^{2} -12x+13[/tex]

Learn more about the composite function here:

https://brainly.com/question/10687170

Answer:

9x² + 28x - 32

Step-by-step explanation:

Given

(9x - 8)(x + 4)

Each term in the second factor is multiplied by each term in the first factor, that is

9x(x + 4) - 8(x + 4) ← distribute both parenthesis

= 9x² + 36x - 8x - 32 ← collect like terms

= 9x² + 28x - 32

Answer:

Option D

Step-by-step explanation:

Answer:

So we have [tex]\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}[/tex].

Step-by-step explanation:

Ok so we are in quadrant 2, that means sine is positive while cosine is negative.

We are given [tex]\cos(\theta)=\frac{-2}{3}(\frac{\text{adjacent}}{\text{hypotenuse}})[/tex].

So to find the opposite we will just use the Pythagorean Theorem.

[tex]a^2+b^2=c^2[/tex]

[tex](2)^2+b^2=(3)^2[/tex]

[tex]4+b^2=9[/tex]

[tex]b^2=5[/tex]

[tex]b=\sqrt{5}[/tex]  This is the opposite side.

Now to find [tex]\csc(\theta)[/tex] and [tex]\tan(\theta)[/tex].

[tex]\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3}{\sqrt{5}}[/tex].

Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).

So [tex]\csc(\theta)=\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}[/tex].

And now [tex]\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{-2}=\frac{-\sqrt{5}}{2}[/tex].

So we have [tex]\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}[/tex].

Answer:

[tex]tan\theta{3}=-\frac{\sqrt5}{2}[/tex]

[tex]cosec\theta=\frac{3}{\sqrt5}[/tex]

Step-by-step explanation:

We are given that [tex]\theta[/tex] be an  angle in quadrant II and [tex]cos\theta=-\frac{2}{3}[/tex]

We have to find the exact values of [tex]cosec\theta[/tex] and [tex]tan\theta[/tex].

[tex]sec\theta=\frac{1}{cos\theta}[/tex]

Then substitute the value of cos theta and we get

[tex]sec\theta=\frac{1}{-\frac{2}{3}}[/tex]

[tex]sec\theta=-\frac{3}{2}[/tex]

Now, [tex]1+tan^2\theta=sec^2\theta[/tex]

[tex]tan^2\theta=sec^2\theta-1[/tex]

Substitute the value of sec theta then we get

[tex]tan^2\theta= (-\frac{3}{2})^2-1[/tex]

[tex]tan^2\theta=\frac{9}{4}-1=\frac{9-4}{4}=\frac{5}{4}[/tex]

[tex]tan\theta=\sqrt{\frac{5}{4}}=-\frac{\sqrt5}{2}[/tex]

Because[tex] tan\theta [/tex] in quadrant II is negative.

[tex]sin^2\theta=1-cos^2\theta[/tex]

[tex]sin^2\theta=1-(\farc{-2}{3})^2[/tex]

[tex]sin^2\theta=1-\frac{4}{9}[/tex]

[tex]sin^2\theta=\frac{9-4}{9}=\frac{5}{9}[/tex]

[tex]sin\theta=\sqrt{\frac{5}{9}}[/tex]

[tex]sin\theta=\frac{\sqrt5}{3}[/tex]

Because in quadrant II [tex]sin\theta[/tex] is positive.

[tex]cosec\theta=\frac{1}{sin\theta}=\frac{1}{\frac{\sqrt5}{3}}[/tex]

[tex]cosec\theta=\frac{3}{\sqrt5}[/tex]

[tex]cosec\theta[/tex] is positive in II quadrant.

Answer:

Two line segments are congruent if they have the same length. Two angles are congruent if they have the same measure.

Two angles are said to be congruent if they are equal. For example, if two triangles each have an angle of 42 degrees, then those angles are congruent.

Find the area of the rug by multiplying the length by the width:

7 x 5 = 35 square meters.

Now multiply the area of the rug by the cost:

35 square meters x 285 pesos per square meter = 9,975 total pesos.

Answer:

2, 3, 4, 5, 6, 7 or 8.

Step-by-step explanation:

We know that the sum of two sides on a triangle should ALWAYS be greater than the third side. Then we have:

5-4 < x < 5 + 4

1 < x < 9

Therefore, the lenght of the third side could be any number between 1 and 9. If the lenght of the third side is an integrer, then the lenght could be:

2, 3, 4, 5, 6, 7 or 8.

Answer:

The length of the third side could be all real numbers greater than 1 unit and less than 9 units

Step-by-step explanation:

we know that

The Triangle Inequality Theorem,   states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

so

Applying the triangle inequality theorem

Let

x ----> the length of the third side

1) 4+5 > x

9 > x

Rewrite

x < 9 units

2) 4+x > 5

x > 5-4

x > 1 units

therefore

The solution for the third side is the interval -----> (1,9)

All real numbers greater than 1 unit and less than 9 units

therefore

The length of the third side could be all real numbers greater than 1 unit and less than 9 units

Answer:

The actual distance between the cities is 480 miles.

Step-by-step explanation:

2.5 inches is 20/8 inches, multiply 20 by 24 miles which gives you 480.

Answer:

480 miles.

Step-by-step explanation:

By proportion the actual distance is

(2.5 / 1/8) * 24

= (2.5 / 0.125) * 24

= 20 * 24

= 480 miles.

The zeros of the function are 2 and -1.

The zeros of a function are the values of x when f(x) is equal to 0.

Given function is,  [tex]f(x)=\frac{(x-2)(x+1)}{x(x-3)(x+5)}[/tex]

Equate given function to zero.

          [tex]\frac{(x-2)(x+1)}{x(x-3)(x+5)} =0\\\\(x-2)(x+1)=0\\\\x=2,x=-1[/tex]

Learn more about the Zeros of function here:

https://brainly.com/question/446160

Answer:

y - 1 = -2(x + 1).

Step-by-step explanation:

The slope = (-3-1)/(1 - -1)

= -4 / 2

= -2.

In point slope form:

y - y1 = m(x - x1)

Using m = -2 and the point (-1, 1):

y - 1 = -2(x + 1).

Answer:

see explanation

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 1, 1) and (x₂, y₂ ) = (1, - 3)

m = [tex]\frac{-3-1}{1+1}[/tex] = [tex]\frac{-4}{2}[/tex] = - 2

Using either of the 2 points as a point on the line, then

Using (- 1, 1)

y - 1 = - 2(x - (- 1)), that is

y - 1 = - 2(x + 1) ← in point- slope form

Answer:

The correct answer is option a.  400

Step-by-step explanation:

Points to remember

Area of square = a²

Where 'a' is the side length of square

To find the area of square

It is given that, the side length of square is 20 inches.

Here a = 20 inches

Area = a²  

 = 20²

 = 400

Therefore the correct answer is option a.  400

Answer:

using distance formula or graphing you can find your answer

Step-by-step explanation:

(7,0) from 4 to -3 is -7. Take the absolute vale of -7 and you get 7.

Hope i helped :)

Answer:

Distance from A to B = 7 units

Step-by-step explanation:

We are given the following two points and we are to find the distance between them:

A (4,2)

B (-3,2)

We will be using the distance formula for this:

Distance = [tex] \sqrt { ( x _ 2 - x _1)^ 2 + ( y _ 2 - y _ 1 ) ^ 2 } [/tex]

AB = [tex]\sqrt{(-3-4)^2+(2-2)^2} = \sqrt{49}[/tex] = 7 units

Answer:

Step-by-step explanation:

For this case we must graph the following function:

[tex]f (x) = 13x-1[/tex]

We found the cut points:

Cutting point with the y-axis:

We make [tex]x = 0,[/tex]

[tex]y = 13 (0) -1\\y = 0-1\\y = -1[/tex]

Cutting point with the x axis:

[tex]0 = 13x-1\\1 = 13x\\x = \frac {1} {13} = 0.078[/tex]

It is observed that the line has a positive slope, [tex]m = 13[/tex]

The graph is seen in the attached image.

Answer:

See attached image

Answer:

[tex]5+\sqrt{x+4}[/tex]

Step-by-step explanation:

The conjugate of the radical expression [tex]a+\sqrt{b}[/tex] is  [tex]a-\sqrt{b}[/tex]

The conjugate of the radical expression [tex]a-\sqrt{b}[/tex] is  [tex]a+\sqrt{b}[/tex]

The sign of the radical becomes its additive inverse in the conjugate,

The given expression is

[tex]5-\sqrt{x+4}[/tex] where [tex]x>-4[/tex] (domain)

The conjugate of this expression is [tex]5+\sqrt{x+4}[/tex]

Answer:

The conjugate is 5+√x+4

Step-by-step explanation:

To find the conjugate of of expression 5-√x+4

when x>-4

First let us understand that in simple terms terms the conjugate of a radical simply involves the change in sign of the radical

in the problem the conjugate of

5-√x+4 is 5+√x+4

it is that simple Just alternate the sign and you are done!!!