Point B is on line segment AC. Given AC = 20 and BC = 16, determine the length AB

Answer:

a = -2

Step-by-step explanation:

Given that,

-1 is a zero of the polynomial. i.e. x = -1.

The polynomial is :

p(x) = ax²-x²+x+4

We need to find the value of a.

Put x = -1 in the above polynomial.

p(1) = a(-1)²-(-1)²+(-1)+4

For zeroes,

p(1) = 0

a(-1)²-(-1)²+(-1)+4 = 0

a-1-1+4 = 0

a = -2

So, the value of a is equal to -2.

Answer:

5.97

Step-by-step explanation:

Answer:

5.97

Step-by-step explanation:

5.946 = 5.94

5.97>5.94

Answer:

[tex]\frac{1}{12}[/tex]

Step-by-step explanation:

We require the fractions to have the same common denominator

The LCM of 4 and 3 is 12 , then

[tex]\frac{3(3)}{4(3)}[/tex] - [tex]\frac{2(4)}{3(4)}[/tex]

= [tex]\frac{9}{12}[/tex] - [tex]\frac{8}{12}[/tex]

= [tex]\frac{1}{12}[/tex]

Answer:

Of is a keyword for multiplication.

So, your equation is really 2 X 1,800,000÷3=?

Final answer: 1,200,000

Step-by-step explanation:

Hope that this helps! :)

Have a great rest of your day/night!

Answer:

[tex]Q_3 = 56.45[/tex] --- The third quartile

[tex]Var = 370.18[/tex] -- Variance

Step-by-step explanation:

Given

[tex]Q_1 = 30.8[/tex] -- First quartile

[tex]Q_2 = 48.5[/tex] --- Median

[tex]\bar x = 42[/tex] --- Mean

[tex]Skp = -0.38[/tex] --- Coefficient of skewness

Solving (9): The third quartile [tex]Q_3[/tex]

This is calculated from

[tex]Skp = \frac{Q_1 + Q_3 - 2Q_2}{Q_3 - Q_1}[/tex]

So, we have:

[tex]-0.38 = \frac{30.8 + Q_3- 2*48.5}{Q_3 - 30.8}[/tex]

Cross Multiply

[tex]-0.38 (Q_3 - 30.8)= 30.8 + Q_3- 2*48.5[/tex]

Open bracket

[tex]-0.38Q_3 + 11.704= 30.8 + Q_3- 97.0[/tex]

Collect like terms

[tex]-0.38Q_3 -Q_3= 30.8 - 97.0- 11.704[/tex]

[tex]-1.38Q_3= -77.904[/tex]

Divide both sides -1.38

[tex]Q_3 = \frac{-77.904}{-1.38}[/tex]

[tex]Q_3 = 56.45[/tex] --- approximated

Solving (b): The variance

First, calculate the standard deviation from:

[tex]3IQR = 4SD[/tex]

[tex]IQR= Q_3 - Q_1[/tex]

So:

[tex]3IQR = 4SD[/tex]

[tex]3(Q_3 - Q_1) = 4SD[/tex]

Make SD the subject

[tex]SD = \frac{3}{4}(Q_3 - Q_1)[/tex]

[tex]SD = \frac{3}{4}(56.45 - 30.8)[/tex]

[tex]SD = \frac{3}{4}*25.65[/tex]

[tex]SD = \frac{3*25.65}{4}[/tex]

[tex]SD = \frac{76.95}{4}[/tex]

[tex]SD = 19.24[/tex]

So, the variance is:

[tex]Var = SD^2[/tex]

[tex]Var = 19.24^2[/tex]

[tex]Var = 370.18[/tex]

Answer:

Always

Assumption:

At least how I know it.

Answer:

cot(x)=-sqrt(3)/3

Step-by-step explanation:

I'm going to use the Pythagorean Identity sin^2(x)+cos^2(x)=1 to find the sine value since cotangent value is the cosine value over sine value.

Plug in:

sin^2(x)+(-1/2)^2=1

Simplify:

sin^2(x)+1/4=1

Subtract 1/4 on both sidea:

sin^2(x)=3/4

Square root both sides:

sin(x)=sqrt(3)/2. [Sine is positive since x is in 2nd quadrant]

cot(x)=(-1/2)/(sqrt(3)/2))

cot(x)=-1/sqrt(3)

Rationalize denominator by multiplying by sqrt(3)/sqrt(3):

cot(x)=-sqrt(3)/3

cot(x)=

Answer:

y = (1/2)x - 2

Step-by-step explanation:

Slope intercept form: y = mx + c

x - 2y = 4

    -2y = -x + 4

[tex]\frac{-2y}{-2}=\frac{-x}{-2}+\frac{4}{-2}\\\\y = \frac{1}{2}x - 2[/tex]

Answer:

b = 6 sqrt(2)

Step-by-step explanation:

Since this is right triangle, we can use the Pythagorean theorem

a^2+b^2 = c^2  where a and b are the legs and c is the hypotenuse

3^2 + b^2 = 9^2

9+b^2 = 81

b^2 = 81-9

b^2 = 72

Take the square root of each side

sqrt(b^2) = sqrt(72)

b = sqrt(36*2)

b = 6 sqrt(2)

The formula for the circumference of a circle is 2 x 3.14 x radius

2 x 3.14 x 6 = 37.68

Answer:

4f + 3g

Step-by-step explanation:

g + 3f + f + g + g

3f + f = 4f

g + g = 2g + g = 3g

4f + 3g

Answer:

Step-by-step explanation:

When a limit has the x value approaching infinity, we are talking about end behavior of functions which implies horizontal asymptotes. As x approaches both positive and negative infinity, the tails of each branch of this function approaches 0; therefore, the correct choice for this the third one down.

Answer: c

Step-by-step explanation: got it right

Step-by-step explanation:

X-2y=5 ; 2x-4y=10

• x - 2y = 5 • 2x -4y = 10

-2y = -x+5 -4y= -2x +10

y1 = ½x - 5/2 y2 = ½x -5/2

• => y1 = y2

½x -5/2 = ½x -5/2

the systems have no solution

• graph as attached

Answer:

[tex]\textsf{Equation 1}: \quad y=\dfrac{1}{2}x-\dfrac{5}{2}[/tex]

[tex]\textsf{Equation 2}: \quad y=\dfrac{1}{2}x-\dfrac{5}{2}[/tex]

True for all x (infinitely many solutions)

Step-by-step explanation:

Slope-intercept form of a linear equation:

[tex]y = mx + b[/tex]

(where m is the slope and b is the y-intercept)

Given equations:

[tex]\begin{cases}x-2y=5\\2x-4y=10\end{cases}[/tex]

To put both equations in slope-intercept form, rearrange both equations to make y the subject:

Equation 1

[tex]\implies x-2y=5[/tex]

[tex]\implies x-2y+2y=5+2y[/tex]

[tex]\implies x=5+2y[/tex]

[tex]\implies x-5=5+2y-5[/tex]

[tex]\implies 2y=x-5[/tex]

[tex]\implies 2y \div 2=(x-5) \div 2[/tex]

[tex]\implies y=\dfrac{1}{2}x-\dfrac{5}{2}[/tex]

Equation 2

[tex]\implies 2x-4y=10[/tex]

[tex]\implies 2x-4y+4y=10+4y[/tex]

[tex]\implies 2x=10+4y[/tex]

[tex]\implies 2x-10=10+4y-10[/tex]

[tex]\implies 4y=2x-10[/tex]

[tex]\implies 4y \div 4=(2x-10) \div 4[/tex]

[tex]\implies y=\dfrac{1}{2}x-\dfrac{5}{2}[/tex]

Therefore, both equations are the same as they have the same slope and y-intercept.

As both equations are the same, this system of equations is called an equivalent system.

To solve a system of equations by substitution, substitute one equation into the other:

[tex]\implies \dfrac{1}{2}x-\dfrac{5}{2}= \dfrac{1}{2}x-\dfrac{5}{2}[/tex]

As both sides are equal, the solution is true for all x.

The solution to a graphed system of equations is the point(s) of intersection.

When graphing the given system of equations, as the lines are the same, they will be on top of each other. They are called coincident lines and have infinitely many solutions since the lines have the same slope and same y-intercept.

Learn more about systems of equations here:

https://brainly.com/question/21447811

https://brainly.com/question/27357423

The answer:

(x+2)²+(y–1)²=9

Answer:

1. 5.75 * 33 = $189.75

The bakers income is $189.75

2. 2 * y - 8 / 4 when y = 7

  2 * 7 - 8 / 4 = 12

3. 76 + 4 * y^2 + 1 when y = 3

   76 + 4 * 3^2 + 1 = 113

======================================================

Explanation:

Think of 25 as 25/1 and 5 as 5/1

Choice E is really saying [tex]\frac{25}{1} \div \frac{5}{1}[/tex] which becomes [tex]\frac{25}{1} \times \frac{1}{5}[/tex]. Note the second fraction flips and we change to a multiplication sign.

------------

Or you could think of it like this:

[tex]25 \times \frac{1}{5} = \frac{25}{1} \times \frac{1}{5} = \frac{25*1}{1*5} = \frac{25}{5} = 25 \div 5[/tex]

to help see why the answer is E.

Answer:

E

Step-by-step explanation:

Answer:

answer is 1

Step-by-step explanation:

3 and 3 gets cancel .when you divide the same number answer is always 1

Answer:

f(x) = - 2(x - 3)² + 2

Step-by-step explanation:

Given

f(x) = - 2x² + 12x - 16 ( factor out - 2 from the first 2 terms )

      = - 2(x² - 6x) - 16

To complete the square

add/subtract ( half the coefficient of the x- term)² to x² - 6x

f(x) = - 2(x² + 2(- 3)x + 9 - 9) - 16

     = - 2(x - 3)² + 18 - 16

     = - 2(x - 3)² + 2

Answer:

Has a slope of [tex]\frac{3}{4}[/tex] and passes through the point (0,1)

Step-by-step explanation:

We can eliminate half the answer choices by finding whether the line passes through (1,0) or (0,1). Coordinates are read as (x,y), therefore the line passes through (0,1). The slope of the line y=x has a slope of 1. Looking at the graph, we can see that the increasing interval of the line is tilted down ever so slightly, therefore the slope must be less than one.

Therefore, the answer is has a slope of [tex]\frac{3}{4}[/tex] and passes through the point (0,1).

Answer:

Its C, or the third option.

Step-by-step explanation:

Rise/run on this graph is 3/4

and y-intercept is (0, 1)

so C, or third option is correct.

Answer:

lincth times with times hight

Step-by-step explanation:

Answer:

I think the answer is 11 because 244140625/(5^3*5^(-2)*5^11)

m=11

[tex] \huge \bf\pink{a} \purple{n} \pink{s} \purple{w} \pink{e} \purple{r}[/tex]

[tex]5m = 60 \\ m = \frac{60}{5} = 12[/tex]

Answer:

AB ≈ 2.62 cm

Step-by-step explanation:

The arc AB is calculated as

AB = circumference of circle × fraction of circle

     = 2πr × [tex]\frac{15}{360}[/tex]

     = 2π × 10 × [tex]\frac{1}{24}[/tex]

     = [tex]\frac{20\pi }{24}[/tex]

     ≈ 2.62 cm ( to the nearest hundredth )

Answer:

[tex]AB\approx 24.22[/tex]

Step-by-step explanation:

In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse of the triangle.

In this case, in order to form the ratio, we're looking for the angle to the bottom right of the triangle. Notice how this angle is the supplement to the angle marked [tex]130.8^{\circ}[/tex]. Since the two angles form one side of the line, we can solve for this angle.

Let's call the angle [tex]\theta[/tex]:

[tex]130.8+\theta=180,\\\theta=49.2^{\circ}[/tex]

From basic. trig for right triangles (refer to first paragraph in answer):

[tex]\sin 49.2^{\circ}=\frac{AB}{32},\\AB=32\sin 49.2^{\circ}=24.2238417809\approx \boxed{24.22}[/tex]

Answer:

100+0.30p

hope this helps

have a good day :)

Step-by-step explanation:

Answer:

a) (-1,4.5)

b)10.63 units

c)7/8

Step-by-step explanation:

a) To find this, we use the midpoint formula

(x,y) = (x1 + x2)/2 , (y1 + y2)/2

(x1,y1) = (3,8)

(x2,y2) = (-5,1)

(x,y) = (3 -5)/2 , (8 + 1)/2 = (-1, 4.5)

b) To calculate the distance D between two points, we use the formula;

D = √(x2-x1)^2 + (y2-y1)^2

D = √(-5-3)^2 + (1-8)^2

D = √(-8)^2 + (-7)^2

D = √64 + 49

D = √113

D = 10.63 units

C) We use the slope formula here

m = (y2-y1)/(x2-x1)

m = (1-8)/(-5-3) = -7/-8 = 7/8

Answer:.

Answer:1.we have dividend =divisor*quotient+remainder=

(x²+2x-1)(x²-2x+1)+(-4x+1)

=x²(x²-2x+1)+2x(x²-2x+1)-1(x²-2x+1)-4x+1

=x⁴-2x³+x²+2x³-4x²+2x-x²+2x-1-4x+1

=x⁴-4x²

Dividend is x⁴-4x².

2.

Multiplier is a-a²+1.

Answer:

[tex]i \: hope \: the \: above \: picture \\ helps[/tex]

[tex]have \: a \: nice \: day \: [/tex]

[tex]#Liliflim[/tex]