The functions f(x) and g(x) are graphed.
f(x)
324
2
-6-5-4-3-2-1₁.
297 49
-2+
-3
-4
g(x)
1 2 3 4 5 6 x
Which represents where f(x) = g(x)?
Of(0) = g(0) and f(2)= g(2)
Of(2)= g(0) and f(0) = g(4)
Of(2)= g(0) and f(4) = g(2)
Of(2)= g(4) and f(1) = g(1)

Answer:

£13.75

Step-by-step explanation:

→ Call the charge x

x

→ Multiply by 16

16x

→ Equate to total

16x = 220

→ Divide both sides by 16

x = £13.75

Answer:$225,000 each

Step-by-step explanation: Assuming that they each get an equal share of the inheritance, there are 8 shares of the inheritance so we divided the $1.8m by 8 which equals $225,000 each.

[tex]m = \frac{ - 1}{a} = \frac{ - 1}{2} \\ y = \frac{ - 1}{2} x + b[/tex]

Answer:

5/4

Step-by-step explanation:

(5/8)(2/1)

(Reciprocal of 1/2 is 2/1 as shown above)

(5)(2)/(8)(1)=10/8

10/8=5/4

Answer:

Option D is correct.

No solution

Step-by-step explanation:

2(x+7)=8

Apply 2 into bracket:

2x+14=8

2x=8-14

2x=-6

Divide both sides by 2:

x=-3

hence here is no option for x=-3, so answer is no solution.

The measure of the angle <JKL is 56 degrees

The theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

Given the following parameters

m<IL = 112 degrees

Since the measure of the vertex is half the measure of its intercepted arc, hence;

<JKl = 1/2(m<IL)

Substitute the given parameters into the formula to have;

<JKl = 1/2(112)

<JKL = 56 degrees

Hence the measure of the angle <JKL is 56 degrees

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On dividing we get 69/2 or 34.5 as remainder.

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem. The factor theorem states that a polynomial f(x) has a factor if and only if f=0.

Long division is best suited for such expressions:-

On applying long division and factor theorem we get

2x²+7x-39 = (2x - 1/2)(x-7) + 69/2

Thus on dividing we get 69/2 or 34.5 as remainder.

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Answer:[tex]\Large\boxed{x=12~;~y=18}[/tex]

Step-by-step explanation:

Given a system of equations

[tex]1)~\frac{x}{y} =\frac{2}{3}[/tex]

[tex]2)~\frac{y}{24} =\frac{3}{4}[/tex]

Multiply 24 on both sides of 2) equation to isolate y-value

[tex]\frac{y}{24}\times24 =\frac{3}{4}\times24[/tex]

[tex]\Large\boxed{y=18}[/tex]

Substitute the y-value into the 1) equation

[tex]\frac{x}{18} =\frac{2}{3}[/tex]

Multiply 18 on both sides of 1) equation to isolate y-value

[tex]\frac{x}{18}\times18 =\frac{2}{3}\times18[/tex]

[tex]\Large\boxed{x=12}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

For direct variation, variation constant = 2.

For inverse square variation, variation constant = 4.

For inverse variation, variation constant =80.

For inverse variation, variation constant = 9.

For direct square variation, variation constant = 0.6

For direct variation, variation constant = -0.84

(a) Direct variation: y = Kx where x = 3 and y = 6

Variation constant(k) = y/x = 6/3 =2

(b) Inverse square variation: y = K/[tex]x^{2}[/tex] where x = -2 and y = 1

Variation constant(k) = y[tex]x^{2}[/tex] = 1(-2)(-2) =4

(c) Inverse variation: y = K/x where x = 8 and y = 10

Variation constant(k) = y*x = 8*10 =80

(d) Inverse variation: y = K/x where x = 3 and y = 3

Variation constant(k) = y*x = 3*3 =9

(e) Direct square variation: y = K[tex]x^{2}[/tex] where x = 5 and y = 15

Variation constant(k) = y/[tex]x^{2}[/tex] = 15/25 =0.6

(f) Direct variation: y = Kx where x = -7.5 and y = 6.3

Variation constant(k) = y/x = -6.3/7.5 = -0.84

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The points if the slope is 4 and the point is (1,-1) are (1,-1) and (2, 3)

Linear equations are equations that have constant average rates of change, slope or gradient

The given parameters are:

Slope = 4

Point = (1, -1)

The slope of 4 means that as x changes by 1, the value of y changes by 4

This means that, we have:

(x + 1, y + 4)

Substitute the known values in the above equation

(1 + 1, -1 + 4)

Evaluate

(2, 3)

Hence, the points if the slope is 4 and the point is (1,-1) are (1,-1) and (2, 3)

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MOV EBX,EAX

SHL EAX,4

SHL EBX,1

ADD EAX,EBX these are the  series of instructions that will multiply eax by 18, using a combination of shift, mov, and add instructions.

STEPS:

1. Copy EAX into another register, say EBX

2. We know that shift left by 'n' bits results in multiplication with 2^'n' , so perform shift left operation of EAX by 4 bits ( results in multiplication of EAX with 16) and perform shift left operation of EBX by 2 bits(results in multiplication of EBX with 2)

3. Now add EAX and EBX which is the required answer ( Since, i*16 + i*2 equals to i*18 )

INSTRUCTIONS:

mov ebx,eax ; make copy

shl eax,4 ; eax * 16

shl ebx,1 ; ebx * 2

add eax,ebx ; answer

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Answer:

Jamie is currently 15.

Step-by-step explanation:

Answer: 11 units

Step-by-step explanation:

We can say that [tex]\triangle TSQ\sim\triangle PSR[/tex] by AA Similarity Postulate. This is since [tex]\angle T\cong\angle P[/tex] and both ∠TSQ and ∠RSP are right angles, making them congruent.

Similar triangles have a property that corresponding sides are proportional. Hence, we can say that

[tex]\frac{ST}{PS}=\frac{SQ}{SR}\\\frac{15}{PS}=\frac{9}{12}\\\frac{15}{PS}=\frac{3}{4}\\60=3*PS\\PS=20[/tex]

We also know that PS is the combined length of PQ and QS. Since we know that QS is 9, let's substitute PQ + 9 in and solve.

[tex]PQ+9=20\\PQ=11[/tex]

The Distance - Time graph that represents that represents Charlie's trip is attached accordingly.

A distance-time graph can be used to illustrate the distance traveled by an item moving in a straight line.

The gradient of the line in a distance-time graph equals the speed of the item. The quicker the item moves, the higher the gradient (and the steeper the line).

Some of the uses of the distance-time graph are;

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Answer:

127

Step-by-step explanation:

Answer:

b

Step-by-step explanation:

The minimum value of f(4) - f(1) is 6.

The maximum value of f(4) - f(1) is 12.

In the question, we are given that, 2 ≤ f'(x) ≤ 4 for all values of x.

Taking the given inequality as (i).

We are asked to find the minimum and maximum possible values of f(4) - f(1).

We multiply (i) by dx throughout, to get:

4dx ≤ f'(x)dx ≤ 5dx.

To find this, we integrate (i) in the definite interval [4, 1] with respect to dx, to get:

[tex]\int_{1}^{4}2dx \leq \int_{1}^{4}f'(x)dx \leq \int_{1}^{4}4dx\\\Rightarrow [2x]_{1}^{4} \leq [f(x)]_{1}^{4} \leq [4x]_{1}^{4}\\\Rightarrow 2*4 - 2*1 \leq f(4)-f(1) \leq 4*4 - 4*1\\\Rightarrow 6 \leq f(4) -f(1) \leq 12[/tex]

Thus, the minimum value of f(4) - f(1) is 6.

The maximum value of f(4) - f(1) is 12.

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Answer:

2

Step-by-step explanation:

The trick here is to reduce each of the bases to lowest form first

[tex]12 = 3.4 = 3 .2.2 = 3.2^{2}\\12^{3} = (3.2^{2} )^{3} = 3^{3}2^{6} \\6 = 3.2\\6^{5} = (3.2)^{5} = 3^{5} .2^{5} \\\\Numerator = 3^32^63^52^5 = 3^82^{11}\\Denominator computation\\9 = 3^2\\9^{4} = (3^{2} )^4 = 3^{8} \\Denominator = 3^82^{10} \\\\\frac{Numerator}{Denominator } =\frac{{3^8}{2^{11} }}{3^82^{10}} = \frac{2^{11}} {2^{10}} = 2^1 = 2[/tex]  

The solution accurate to equation within [tex]10^{-2}[/tex] for [tex]x^{4}-2x^{3} -4x^{2} +4x+4=0[/tex] lies in [0,2].  

Given the equation  [tex]x^{4}-2x^{3} -4x^{2} +4x+4=0[/tex] and range is  [tex]10^{-2}[/tex].

We are required to find the interval in which the solution lies.

The attached table shows the iterations. At each step, the interval containing the root is bisected and the function value at the mid point of the interval is found. The sign of its relative to the signs of the function values at the ends of the interval tell which half interval contains the root. The process is repeated until the interval width is less than [tex]10^{-2}[/tex].

Interval:[0,2], signs [+,-],mid point:1, sign at midpoint +.

[1,2]                                                       3/2

[1,3/2]                                                    5/4

The rest is in the attachment. The listed table values are the successive interval mid points.

The final midpoint is 181/128=1.411406.

This solution is within 0.0002 of the actual root.      

Hence the solution accurate to equation within [tex]10^{-2}[/tex] for [tex]x^{4}-2x^{3} -4x^{2} +4x+4=0[/tex] lies in [0,2].

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Answer:

9

Step-by-step explanation:

let the number be y

5×y - 7 = 3×y + 11

5y-7 = 3y+11

collect like terms

5y-3y = 11+7

2y = 18

y = 18/2

y = 9

Answer:

15^6

Step-by-step explanation:

According to BODMAS

where;

B - Bracket ()

O - Off (*)

D - Division (/)

M - Multiplication (*)

A - Addition (-)

S - Subtraction (+)

Bracket should be solved first

therefore;

(3*5)^6

15^6 = 11390625

Which can be approximately in standard form written as 1.1*10^7

Answer:

1.09575e+11

Step-by-step explanation:

There are only 13 multiples of both 3 and 5, since

[tex]200 = 3\cdot5\cdot13 + 5[/tex]

From these integers, we eliminate any that are also divisible by 4 or 7.

[tex]3\cdot5\cdot4 = 60 \implies \{60,120,180\}[/tex]

[tex]3\cdot5\cdot7 = 105 \implies \{105\}[/tex]

so there are 13 - 4 = 9 such integers.

Answer:

Step-by-step explanation:

The measurement used to describe a television is the length of its diagonal. The relation between the width, height, and diagonal is described by the Pythagorean theorem.

The Pythagorean theorem tells us the relation between the sides of a right triangle and its hypotenuse. The diagonal of a rectangle is the hypotenuse of a right triangle whose sides are the width and height of the rectangle. If 'c' is the number of "ratio units" in the diagonal, we have ...

  4² +3² = c²

  c = √(16 +9) = 5

The diagonal of the screen is 5 ratio units, so the width is 4/5 of the length of the diagonal, and the height is 3/5 the length of the diagonal.

The width is ...

  (4/5)(47 in) = 37.6 in . . . width

The height is ...

  (3/5)(47 in) = 28.2 in . . . height

The area is the product of the width and height:

  A = WH = (37.6 in)(28.2 in) = 1060.32 in²

The area of the screen is 1060.32 square inches.

Answer:

b 19.2

Step-by-step explanation:

a = [tex]\pi[/tex][tex]r^{2}[/tex] for a circle.  We do not want to find the area for a whole circle.  We only want to find the area for part of a circle.  a hole circle is 360 degrees.

a = [tex]\frac{88}{360}[/tex][tex]\pi[/tex][tex]r^{2}[/tex]

a = [tex]\frac{88}{360}[/tex][tex]\pi[/tex]([tex]5^{2}[/tex])

a = 19.2 rounded.

The area of the sector of circle is b. 19.20 square meter.

The space enclosed by the sector of circle is called area of the sector of circle.Mathematically,

Area of the sector of circle, A = θ/360πr²

where θ is the angle of the arc and r is the radius of the circle.

Now it is given that,

radius of the circle, r = 5m

Angle of arc, θ = 88°

Therefore, area ofsector of circle A = θ/360πr²

Put the values,

A = 88/360 π 5²

Solving the equation we get

A = 19.20 square meter.

Hence,the area of the sector of circle is b. 19.20 square meter.

So  the correct answer is b.) 19.20 square meter.

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a) The coordinates for T and H such that quadrillateral MATH is a rectangle are T(x, y) = (2, - 2) and H(x, y) = (5, - 4).

b) The quadrilateral MATH is a rectangle.

Quadrilaterals are figures with four sides and whose internal angles sums a total of 360 degrees. A quadrilateral is a rectangle when each pair of opposite sides are parallel and have the same length to each others, each of the four angles are right angles and each pair of perpendicular sides.

a) Vectorially speaking, we can construct a rectangle by using the following definitions:

M(x, y) = (a, b), A(x, y) = (c, d), T(x, y) = (a, d), H(x, y) = (c, b)

If we know that a = 2, b = - 4, c = 5, d = - 2, then the points T and H are described below:

T(x, y) = (2, - 2), H(x, y) = (5, - 4)

The coordinates for T and H such that quadrillateral MATH is a rectangle are T(x, y) = (2, - 2) and H(x, y) = (5, - 4).

b) To prove that quadrilateral MATH, we need to prove that:

MT = HA      

(2, - 2) - (2, - 4) = (5, - 2) - (5, - 4)

(0, 2) = (0, 2)

TA = MH

(5, - 2) - (2, - 2) = (5, - 4) - (2, - 4)

(3, 0) = (3, 0)

MT • TA = 0

(0, 2) • (3, 0) = 0 · 3 + 2 · 0 = 0

MH • HA = 0

(3, 0) • (0, 2) = 3 · 0 + 0 · 2 = 0

Therefore, the quadrilateral MATH is a rectangle.

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