Solve the system of equations for x. The value of x is ____. 3x + 4y = −16 x = 4y

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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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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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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Answer:

B

Step-by-step explanation:

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:

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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By the ratio test, the series converges for

[tex]\displaystyle \lim_{k\to\infty} \left|\frac{(x+2)^{k+1}}{\sqrt{k+1}} \cdots \frac{\sqrt k}{(x+2)^k}\right| = |x+2| \lim_{k\to\infty} \sqrt{\frac k{k+1}} = |x+2| < 1[/tex]

so that the radius of convergence is 1, and the interval of convergence is

|x + 2| < 1   ⇒   -1 < x + 2 < 1   ⇒   -3 < x < -1

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

[tex]\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|[/tex] = β

Here [tex]a_{k}[/tex] = [tex]\frac{(x+2)^{k}}{\sqrt{k} }[/tex]  and  [tex]a_{k+1}[/tex] = [tex]\frac{(x+2)^{k+1}}{\sqrt{k+1} }[/tex]

Now limit is taken,

[tex]\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }|[/tex] = [tex]\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|[/tex]

= [tex]\lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|[/tex]

= [tex]\lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|[/tex]

= [tex]|{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}[/tex]

= [tex]|{x+2 }|[/tex] < 1

- 1 < [tex]{x+2 }[/tex] < 1

- 1 - 2 < x < 1 - 2

- 3 < x < - 1

We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

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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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Answer:

[tex]\displaystyle x=\frac{5}{4},\;\;1\frac{1}{4}, \;\; or \;\; 1.25[/tex]

Step-by-step explanation:

    To solve for x, we need to isolate the x variable.

    Given:

[tex]\displaystyle x+\frac{1}{2} =\frac{7}{4}[/tex]

    Subtract [tex]\frac{1}{2}[/tex] from both sides of the equation:

[tex]\displaystyle (x+\frac{1}{2})-\frac{1}{2} =(\frac{7}{4})-\frac{1}{2}[/tex]

[tex]\displaystyle x=\frac{7}{4}-\frac{1}{2}[/tex]

    Now, we will create common denominators to simplify.

[tex]\displaystyle x=\frac{7}{4}-\frac{2}{4}[/tex]

[tex]\displaystyle x=\frac{5}{4}[/tex]

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

3x-2y=  7

Slope intercept form y= mx+b

   3x- 2y=  7

-   3x           -3x

-2y/-2=  3x/-2 - 7/2  

y=   3x/-2 -7/2

Answer:

3x/2 and the second 7/-2y

Step-by-step explanation:

thats how is what i got

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:

3600

Step-by-step explanation:

Reason is because you put the answer in your question.

Answer:

Rs.10800

Step-by-step explanation:

loss of article(L%)=10%

s.p=?

c.p of an article (c.p)=c.p

Now,

s.p=100-L%×c.p

_____

100

= 100-10 ×c.p

_____

100

=90 ×c.p

___

100

=0.9 × c.p

if article sold with Rs. 540 more

Article profit =5%

s.p =c.p + 540 =100+p%×c.p

______

100

or, c.p +540=105×c.p

______

100

or, c.p +540=1.05×c.p

or,c.p +540=1.05c.p

or, 540=1.05c.p-c.p

or,540=0.05c.p

or, 540=c.p

____

0.05

Therefore, c.p =Rs. 10800

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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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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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]

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.

Answer:

Jamie is currently 15.

Step-by-step explanation:

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

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:

130 degrees

Step-by-step explanation:

We know (x-10)+(2x+10) is 180 degrees.

So first we need to find x. (x-10)+(2x+10)=3x

3x=180

x=60

Angle TRV is 2x+10.

If we input 60 in the place of x, it is 2(60)+10=120+10=130.

The answer is 130 degrees.

Answer:

1.09575e+11

Step-by-step explanation:

The equation of a circle exists:

[tex]$(x-h)^2 + (y-k)^2 = r^2[/tex], where (h, k) be the center.

The center of the circle exists at (16, 30).

Let, the equation of a circle exists:

[tex]$(x-h)^2 + (y-k)^2 = r^2[/tex], where (h, k) be the center.

We rewrite the equation and set them equal :

[tex]$(x-h)^2 + (y-k)^2 - r^2 = x^2+y^2- 32x - 60y +1122=0[/tex]

[tex]$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - r^2 = x^2 + y^2 - 32x - 60y +1122 = 0[/tex]

We solve for each coefficient meaning if the term on the LHS contains an x then its coefficient exists exactly as the one on the RHS containing the x or y.

-2hx = -32x

h = -32/-2

⇒ h = 16.

-2ky = -60y

k = -60/-2

⇒ k = 30.

The center of the circle exists at (16, 30).

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