Select the correct answer.
What is the solution to the equation?
2(x + 7) = 8
OA. -15 and 1
OB. -15
OC. 1
OD.
no solution

Answer:

x=2.5

Step-by-step explanation:

Solve for x in this systems of equations, we already have two values which equal y so half the work is finished for us.

Starting with the equation:

[tex]x+12=17-x[/tex]

Add x to both sides to remove the negative x on the right side of the equation:

[tex]2x+12=17[/tex]

Subtract 12 from both sides to remove the 12 on the left side of the equation:

[tex]2x=5[/tex]

Divide both sides by 2 to cancel out the x coefficient

[tex]x=2.5[/tex]

Answer:

109°

Step-by-step explanation:

The sum of angles in all triangles is equal to 180. For this given triangle, we can solve for x.

48° + 23° + x° = 180°
180° - 71° = x°
x = 109°

Answer:

Step-by-step explanation:

A= π(5)²

 = 3.14(5)²

 = 3.14(25)

 = 78.5 ans

Answer:

-456 I think but I could be wrong though

Option ( C ) is correct for this expression . 100 inches Over 1 minute times × 1 foot Over 12 inches.

What is a basic expression?

Given that the expression 100 inches.

We need to convert 100 inches per minute to feet per minute.

Since, we know that 1ft = 12 inch

Then,

1 in = 1/12 ft

Now, we shall convert 100 inches per minute to feet per minute.

To convert in/min to ft/min, let us multiply by 1/12

Thus, we have,

[tex]\frac{100 in}{min} * \frac{1 ft }{12 in}[/tex]

Therefore, option ( c ) is correct for this expression .

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The complete question is -

Which expression converts 100 inches per minute to feet per minute?

A) Start Fraction #1 100 inches Over 1 minute End Fraction × Start Fraction 60 minutes Over 1 hour End Fraction

B) Start Fraction #2 100 inches Over 1 minute End Fraction × Start Fraction 1 hour Over 60 minutes End Fraction

C) 100 inches Over 1 minute times × Start Fraction 1 foot Over 12 inches End Fraction

D) 100 inches Over 1 minute times × Start Fraction 12 inches Over 1 foot End Fraction

Answer:

K(6, 9)

Step-by-step explanation:

Let the K coordinates be (x, y)

Mid-point formula:

[tex]\sf (x_m, y_m) = (\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2})[/tex]

Applying formula:

[tex]\sf (-3,\:-1) = (\:\dfrac{-12+x}{2} ,\: \dfrac{-11+y}{2} )[/tex]

Comparing expression:

[tex]\sf \dfrac{-12+x}{2} = -3 \quad and \quad \dfrac{-11+y}{2} = -1[/tex]

[tex]\sf -12+x = 2(-3) \quad and \quad -11+y = 2(-1)[/tex]

[tex]\sf x = -6 + 12 \quad and \quad y = -2 + 11[/tex]

[tex]\sf x = 6 \quad and \quad y = 9[/tex]

So, coordinates of K is (6, 9)

Answer:

The length of the diagonal of a rectangle is 10m

Step-by-step explanation:

There is a visual below,

The diagonal of a rectangle has formed 2 right triangles.

This diagonal is called hypotenuse side length of a triangle

To find the hypotenuse, we can use the Pythagorean Theorem

a² + b² = c²

we are given the length of a and b, where a = 6 and b = 8

(6)² + (8)² = c²

36 + 64 = c²

100 = c²

√100 = c

10 = c

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[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

let's solve ~

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4}{y - 6} + \cfrac{5}{y + 3} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4(y + 3) + 5(y - 6)}{(y - 6)(y + 3)} = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{4y + 12 + 5y - 30}{ {y}^{2} + 3y - 6y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{9y - 18}{ {y}^{2} - 3y - 18 } = \cfrac{7y - 4}{ {y}^{2} - 3y - 18} [/tex]

[tex]\qquad \sf  \dashrightarrow \: 9y - 18 = 7y - 4[/tex]

[ denominator is same, so numerator must have same value to be equal ]

[tex]\qquad \sf  \dashrightarrow \: 9y - 7y = - 4 + 18[/tex]

[tex]\qquad \sf  \dashrightarrow \: 2y = 14[/tex]

[tex]\qquad \sf  \dashrightarrow \: y = 7[/tex]

Answer:

9/4

Step-by-step explanation:

We follow bodmas

4/3 +( -1/6 + 13/12)

( lcm = 12)

( -2+ 13/12)

( 11/ 12)

4/3 + ( 11/12)

4/3 + 11/12

lcm = 12 also

and that will equal to

=16 + 11/ 12

= 27/ 12

divide by 3 to simplest form

= 9/4

Step-by-step explanation:

3 meters = 300 cm

anyway, the ratio tells us that the whole ribbon can be seen as the combination of 3 + 2 = 5 equally long parts.

one such part is then

3 m / 5 = 0.6 m or 60 cm

the shorter piece is 2 of these parts long :

2 × 0.6 = 1.2 m or 120 cm

a. altitude = CE

b. bisector = BD

c. exterior angle = ∠ABE

d. median = CF

e. remote interior angles = ∠BCE and ∠CEB

From the question, we are to fill in the blanks

In ΔBCE, we have that ∠BCE is a right angle

Thus,

a. altitude = CE

Also, we have that

∠EBD ≅ ∠CBD

Thus, BD is a bisector

b. bisector = BD

The exterior angle of the triangle is ∠ABE

c. exterior angle = ∠ABE

From the given information,

BF ≅ EF

∴ F is the midpoint of BE

NOTE: Median is a line segment joining the vertex of one side of the triangle to the midpoint of its opposite side.

The median of the triangle is CF

d. median = CF

The remote interior angles of the triangle are ∠BCE  and ∠CEB

e. remote interior angles = ∠BCE and ∠CEB

Hence,

a. altitude = CE

b. bisector = BD

c. exterior angle = ∠ABE

d. median = CF

e. remote interior angles = ∠BCE and ∠CEB

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The terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].

In this question,

The function is f(x) = [tex]\frac{sin(x)}{1-x}[/tex]

The general form of Maclaurin series is

[tex]\sum \limits^\infty_{k:0} \frac{f^{k}(0) }{k!}(x-0)^{k} = f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} +\frac{f'''(0)}{3!}x^{3}+......[/tex]

To find the Maclaurin series, let us split the terms as

[tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] ------- (1)

Now, consider f(x) =  sin(x)

Then, the derivatives of f(x) with respect to x, we get

f'(x) = cos(x), f'(0) = 1

f''(x) = -sin(x), f'(0) = 0

f'''(x) = -cos(x), f'(0) = -1

[tex]f^{iv}(x)[/tex] = cos(x), f'(0) = 0

Maclaurin series for sin(x) becomes,

[tex]f(x) = 0 +\frac{1}{1!}x +0+(-\frac{1}{3!} )x^{3} +....[/tex]

⇒ [tex]f(x)=x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+.....[/tex]

Now, consider [tex]f(x) = (1-x)^{-1}[/tex]

Then, the derivatives of f(x) with respect to x, we get

[tex]f'(x) = (1-x)^{-2}, f'(0) = 1[/tex]

[tex]f''(x) = 2(1-x)^{-3}, f''(0) = 2[/tex]

[tex]f'''(x) = 6(1-x)^{-4}, f'''(0) = 6[/tex]

[tex]f^{iv} (x) = 24(1-x)^{-5}, f^{iv}(0) = 24[/tex]

Maclaurin series for (1-x)^-1 becomes,

[tex]f(x) = 1 +\frac{1}{1!}x +\frac{2}{2!}x^{2} +(\frac{6}{3!} )x^{3} +....[/tex]

⇒ [tex]f(x)=1+x+x^{2} +x^{3} +......[/tex]

Thus the Maclaurin series for [tex]f(x)=sin(x)(\frac{1}{1-x} )[/tex] is

⇒ [tex]f(x)=(x-\frac{x^{3} }{3!} +\frac{x^{5} }{5!}+..... )(1+x+x^{2} +x^{3} +......)[/tex]

⇒ [tex]f(x)=x+x^{2} +x^{3} - \frac{x^{3} }{6} +x^{4}-\frac{x^{4} }{6} +.....[/tex]

⇒ [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex]

Hence we can conclude that the terms through degree four of the Maclaurin series is [tex]f(x)=x+x^{2} +\frac{5x^{3} }{6} +\frac{5x^{4} }{6} +.....[/tex].

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The correct equation is (a) because π/1 = C2/r2, [tex]C_2= 2\pi r[/tex]

The complete question is added as an attachment

The given parameters are:

d1 = 1

d2 = 2r

The circumferences of the circles are calculated as:

C = πd

This gives

C1 = π * 1 = π

C2 = π * 2r = 2πr

So, we have:

C1 = π

C2 = 2πr

and

d1 = 1

d2 = 2r

Divide both equations

[tex]\frac{C_2}{d_1} = \frac{C_2}{d_2}[/tex]

[tex]\frac{\pi}{1} = \frac{2\pi r}{2r}[/tex]

This gives

[tex]\frac{\pi}{1} = \frac{C_2}{2r}[/tex]

Hence, the correct equation is (a) because π/1 = C2/r2, [tex]C_2= 2\pi r[/tex]

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Answer:

Its A

Step-by-step explanation:

Answer:

x-int: (4, 0)

Step-by-step explanation:

First, we need to find the equation of the line.

The equation of a line is y = mx + b, where m is the slope and b is the y-intercept.

Thus, we need to find the slope first by using the slope formula:

[tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} } \\[/tex] or change in y / change in x

So we have: [tex]\frac{3-2}{1-2}=\frac{1}{-1}=-1=m[/tex]

We must plug in the slope to find the y-intercept:

[tex]3=-1(1)+b\\3=-1+b\\4=b[/tex]

So the equation of the line is y = -x + 4

The x-intercept means that the y coordinate is 0 so we can use the equation:

[tex]0=-x+4\\-4=-x\\4=x[/tex]

The y-intercept of the linear equation that models the data indicates that:

There was 1 inch of water on the tub when the water was turned on.

A linear function is modeled by:

y = mx + b

In which:

In this problem, when x changes by 1, y changes by 2, hence the slope is of 2. Hence, since when x = 1, y = 3, when x = 0, y = 3 - 2 = 1, which means that the y-intercept is of 1, and the correct interpretation is:

There was 1 inch of water on the tub when the water was turned on.

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The mean, median, and mode all have the same values in a symmetrical distribution. The mean is frequently chosen as the primary indicator of central tendency in these situations.

The mean, median, and mode frequently occur at the same location in a symmetrical distribution, where the values of variables exist at regular frequencies. The graph's middle can be divided into two sides that mirror one another by drawing a line through it.

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Answer:

[tex]\dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{x^2}{\sqrt{25-x^2}}\:\:\text{d}x[/tex]

Rewrite 25 as 5²:

[tex]\implies \displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x[/tex]

Integration by substitution

[tex]\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}[/tex]

[tex]\textsf{Let }x=5 \sin \theta[/tex]

[tex]\begin{aligned}\implies \sqrt{5^2-x^2} & =\sqrt{5^2-(5 \sin \theta)^2}\\ & = \sqrt{25-25 \sin^2 \theta}\\ & = \sqrt{25(1-\sin^2 \theta)}\\ & = \sqrt{25 \cos^2 \theta}\\ & = 5 \cos \theta\end{aligned}[/tex]

Find the derivative of x and rewrite it so that dx is on its own:

[tex]\implies \dfrac{\text{d}x}{\text{d}\theta}=5 \cos \theta[/tex]

[tex]\implies \text{d}x=5 \cos \theta\:\:\text{d}\theta[/tex]

Substitute everything into the original integral:

[tex]\begin{aligned}\displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x & = \int \dfrac{25 \sin^2 \theta}{5 \cos \theta}\:\:5 \cos \theta\:\:\text{d}\theta \\\\ & = \int 25 \sin^2 \theta\end{aligned}[/tex]

Take out the constant:

[tex]\implies \displaystyle 25 \int \sin^2 \theta\:\:\text{d}\theta[/tex]

[tex]\textsf{Use the trigonometric identity}: \quad \cos (2 \theta)=1 - 2 \sin^2 \theta[/tex]

[tex]\implies \displaystyle 25 \int \dfrac{1}{2}(1-\cos 2 \theta)\:\:\text{d}\theta[/tex]

[tex]\implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating $\cos kx$}\\\\$\displaystyle \int \cos kx\:\text{d}x=\dfrac{1}{k} \sin kx\:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\begin{aligned} \implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta & =\dfrac{25}{2}\left[\theta-\dfrac{1}{2} \sin 2\theta \right]\:+\text{C}\\\\ & = \dfrac{25}{2} \theta-\dfrac{25}{4}\sin 2\theta + \text{C}\end{aligned}[/tex]

[tex]\textsf{Use the trigonometric identity}: \quad \sin (2 \theta)= 2 \sin \theta \cos \theta[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{25}{4}(2 \sin \theta \cos \theta) + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{25}{2}\sin \theta \cos \theta + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\sin \theta \cdot 5 \cos \theta + \text{C}[/tex]

[tex]\textsf{Substitute back in } \sin \theta=\dfrac{x}{5} \textsf{ and }5 \cos \theta = \sqrt{25-x^2}:[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\cdot \dfrac{x}{5} \cdot \sqrt{25-x^2} + \text{C}[/tex]

[tex]\implies \dfrac{25}{2} \theta-\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}[/tex]

[tex]\textsf{Substitute back in } \theta=\arcsin \left(\dfrac{x}{5}\right) :[/tex]

[tex]\implies \dfrac{25}{2} \arcsin \left(\dfrac{x}{5}\right) -\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}[/tex]

Take out the common factor 1/2:

[tex]\implies \dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}[/tex]

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The distance the bird has flown by the time Alice and Bob meet is 40 feet.

Given that the distance between Alice and Bob is 1000 feet and their running speed is 10 feet per second and the speed of bird is 20 feet per second.

Distance equals speed multiplied by time.

Distance between Alice and Bob=1000 feet.

Distance between the bird and Bob=1000 feet.

Speed of Alice and Bob=10 feet per second.

The combined speed of Alice and Bob=20 feet per second.

Since the two are running directly toward each other the distance each will cover at the meeting point is 50 feet (1000/20)

The time covered at the meeting point=20 second (1000/50)

Speed of the bird=20 feet per second.

The distance covered by the bird towards Bob at their meeting point is 40 feet(20 feet*20 seconds).

Hence the distance the bird has flown by the time Alice and Bob meet is 40 feet.

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The roots of the given polynomials exists

[tex]$x=8+\sqrt{10},[/tex]  and [tex]$ x=8-\sqrt{10}[/tex]

For a quadratic equation of the form [tex]$a x^{2}+b x+c=0$[/tex] the solutions are

[tex]$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]

Therefore by using the formula we have

[tex]$x^{2}-16 x+54=0[/tex]

Let, a = 1, b = -16 and c = 54

Substitute the values in the above equation, and we get

[tex]$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$[/tex]

simplifying the equation, we get

[tex]$x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1}[/tex]

[tex]$x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1}$[/tex]

[tex]$x=8+\sqrt{10}, x=8-\sqrt{10}[/tex]

Therefore, the roots of the given polynomials are

[tex]$x=8+\sqrt{10},[/tex]  and [tex]$ x=8-\sqrt{10}[/tex].

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The equivalent expression of the radical expression ∛1080 is 6∛5

The radical expression is given as

∛1080

Express 1080 as 216 * 5

∛1080 = ∛(216 * 5)

Split the factors

∛1080 = ∛216 * ∛5

Evaluate the cube root of 216

∛1080 = 6 * ∛5

Evaluate the product

∛1080 = 6∛5

Hence, the equivalent expression of the radical expression ∛1080 is 6∛5

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The solution to the system of inequalities y > 2x + 4 and x + y ≤ 6 on the coordinate plane is the darker region shown in the graph.

An equation is an expression that shows the relationship between two or more variables and numbers.

Inequalities is an expression that shows the non equal comparison of two or more numbers and variables.

The solution to the system of inequalities y > 2x + 4 and x + y ≤ 6 on the coordinate plane is the darker region shown in the graph.

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Answer:

4/10

Step-by-step explanation:

Let the numerator be 2x and the denominator is 5x. The denominator is 6 more than the numerator so if we add 6 to the numerator both numbers should be equal.

Solve:

2x+6 = 5x

6 = 3x

x = 2

2x = 4

5x = 10

Step-by-step explanation:

Let the numerator be 2x and denominator be 5x,

A.T.Q,

2x+6 = 5x

2x-5x = -6

-3x = -6

3x = 6

x = 2

So, the fraction is 4/10 = 2/5

Hello,

[tex]3x + 4 = 8[/tex]

[tex]3x + 4 - 4 = 8 - 4[/tex]

[tex]3x = 4[/tex]

[tex] \frac{3x}{3 } = \frac{4}{3} [/tex]

[tex]x = \frac{4}{3} [/tex]

Answer:

-8

Step-by-step explanation:

Remember pemdas

7 - 3*(10:2):1 =

7-3*5:1=

7-15:1=

7- 15=

-8

Answer:

-8

Step-by-step explanation:

7-3(10:2):1

We always solve what's in parenthesies first.

7 - 3*5:1

Multiplication and division have equal priority, so we solve that part from left to right:

7 - 15:1 =

= 7 - 15 =

= -8

Answer:

Let , the constant factor be x so 3 X + 5 x + 7 x = 180

Step-by-step explanation:

Find out the value of x

7x+3x+5x=180

15x=180

x=180/15

x=12

now,

     x=3 multiplied by 12=36

     x=7 multiplied by 12=84

      x=5 multiplied by 12=60

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Answer:

336

Step-by-step explanation:

They can place 1 of 8 frags on the bottom.

Now they have 7 flags left.

They can place 1 of 7 flags in the middle.

Now they have 6 flags left.

The can place 1 of 6 flags on top.

8 × 7 × 6 = 336

The type of model that best fits the given situation is; A linear equation  Model

Right inside the local reservoir we will have an initial amount of water A.

Now, for every hour that passes by, the amount of water in the reservoir decreases by 500 gals.

Thus, after t hours, the amount of water in the reservoir is expressed as:

W = A - 500gal * t

This is clearly a linear equation model and so we can conclude that the model that fits best in the given situation is a linear model.

The domain of this model is restricted because we can't have a negative amount of water in the reservoir,  and as such the maximum value of t accepted is: W = 0 = A - 500gal*t

t = A/500 hours

Therefore, the domain of this linear relation is: t ∈ {0h, A/500 }

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