quadratic formula 2x^2-9x-35=0

The length of the rectangular book is 5.8cm

In Euclidean plane geometry, a rectangle is a quadrilateral with four right angles. It can also be defined as: an equiangular quadrilateral, since equiangular means that all of its angles are equal (360°/4 = 90°); or a parallelogram containing a right angle. A rectangle with four sides of equal length is a square. The term oblong is occasionally used to refer to a non-square rectangle.

The perimeter of a rectangle is found by adding up the length of all four sides. The formula for the perimeter of a rectangle is,

P = length + breadth + length + breadth. Or, P = 2(length + breadth)

Given:

Perimeter = 16.84 cm.

length = 18/7 cm

substituting in the formula we get

16.84 = 2( 18/7 + width)

8.42 = 18/7 + width

5.8 cm = width

Thus the length of the rectangular book is 5.8cm.

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The linear function h(x) = 2x + 3, with the domain {-2, -1, 0, 1, 2}, has the range {-1, 1, 3, 5, 7}.

To find the range, we substitute each value of the domain for x, in the linear function h(x) = 2x + 3, as follows:

For the domain value, x = -2:

h(-2) = 2(-2) + 3,

or, h(-2) = -4 + 3,

or, h(-2) = -1.

Therefore, for the domain value, x = -2, the range value h(-2) is -1.

For the domain value, x = -1:

h(-1) = 2(-1) + 3,

or, h(-1) = -2 + 3,

or, h(-1) = 1.

Therefore, for the domain value, x = -1, the range value h(-1) is 1.

For the domain value, x = 0:

h(0) = 2(0) + 3,

or, h(0) = 0 + 3,

or, h(0) = 3.

Therefore, for the domain value, x = 0, the range value h(0) is 3.

For the domain value, x = 1:

h(1) = 2(1) + 3,

or, h(1) = 2 + 3,

or, h(1) = 5.

Therefore, for the domain value, x = 1, the range value h(1) is 5.

For the domain value, x = 2:

h(2) = 2(2) + 3,

or, h(2) = 4 + 3,

or, h(2) = 7.

Therefore, for the domain value, x = 2, the range value h(2) is 7.

Therefore, the linear function h(x) = 2x + 3, with the domain {-2, -1, 0, 1, 2}, has the range {-1, 1, 3, 5, 7}.

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Using the z-distribution, the 90% confidence interval for the difference of proportions is given by: (-0.0534, 0.0434).

For each sample, the mean and the standard error are given as follows:

For the distribution of differences, they are given by:

The interval is given by:

[tex]\overline{p} \pm zs[/tex]

In this problem, we have a 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so the critical value is z = 1.645.

Hence the bounds of the interval are:

[tex]\overline{p} - zs = -0.005 - 1.645(0.0294) = -0.0534[/tex]

[tex]\overline{p} + zs = -0.005 + 1.645(0.0294) = 0.0434[/tex]

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The terms are as follows:

t₃ = -62

t₇ = -29.6

tₙ = 10.8n - 105.2

t₁₄ = 46

t₁₉ = 100

Therefore,

tₙ = a + (n - 1)d

where

Hence,

46 = a + 13d

100 = a + 18d

5d = 54

d = 10.8

46 - 13(10.8) = a

a = 46 - 140.4

a = - 94.4

t₃ = a + 3d = - 94.4 + 3(10.8) = -62

t₇ = a + 6d = - 94.4 + 6(10.8) = -29.6

tₙ = -94.4 + 10.8n - 10.8 = 10.8n - 105.2

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Answer:

see attached graphs

Step-by-step explanation:

Transformations

For a > 0

[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]

[tex]f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}[/tex]

[tex]f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}[/tex]

[tex]f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}[/tex]

[tex]y=a\:f(x) \implies f(x) \: \textsf{stretched parallel to the y-axis (vertically) by a factor of}\:a[/tex]

[tex]y=f(ax) \implies f(x) \: \textsf{stretched parallel to the x-axis (horizontally) by a factor of} \: \dfrac{1}{a}[/tex]

[tex]y=-f(x) \implies f(x) \: \textsf{reflected in the} \: x \textsf{-axis}[/tex]

[tex]y=f(-x) \implies f(x) \: \textsf{reflected in the} \: y \textsf{-axis}[/tex]

Parent function:  [tex]f(x) = x[/tex]

Each transformation is a transformation of the parent function only:

Translation of 4 units down

[tex]\implies f(x)-4=x-4[/tex]

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

A stretch by a factor of 3

[tex]\implies 3f(x)=3x[/tex]

[tex]\implies y=3x[/tex]

A reflection and a translation 1 unit up

[tex]\implies f(-x)+1=-x+1[/tex]

[tex]\implies y=-x+1[/tex]

Answer:

1

Step-by-step explanation:

Using a calculator for the chi-square distribution, the critical value is given by:

A) 20.090

For a chi-square critical value, three parameters are needed:

In this problem, we have a one-tailed test(as we are testing if the mean is greater than a value), with a significance level of 0.01 and 8 df, hence, using a calculator, the critical value is of 20.09, hence option A is correct.

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if stretched vertically by a factor of 2, the resulting function will be g(x) = 2|x| and a horizontal compression by a factor of 2, the resulting function will be 1/2|x|

Transformation is a technique used to change the position of an image on an xy-plane.

If the function f(x)is vertically stretched by "a" units, the resulting function will be f(x) = a|x|.

Hence if stretched vertically by a factor of 2, the resulting function will be g(x) = 2|x|

For a horizontal compression by a factor of 2, the resulting function will be 1/2|x|

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The correct option is 'The domain of both functions is all real numbers'.

According to statement

Here, functions are f(x) = 5x and g(x) =5x

Since the domain of f(x) is, D= { x : x belongs to all real numbers.}

Therefore, Domain of f(x) is all real numbers.

Since f(x) and g(x) have the same value,

Therefore, Domain of g(x) is also the set of real numbers.

Now, range of f(x) is , R=  { 5x : x belongs to all real numbers.}

Therefore, Range of f(x) is all real numbers.

Range of g(x) is also all real numbers.

So, we can say by the above explanation, only option (1) is correct.

DISCLAIMER: The question was incomplete. Please find the full content below.

QUESTION

Keshawn is asked to compare and contrast the domain and range for the two functions.

f(x) = 5x

g(x) = 5x

Which statements could he include in his explanation? Check all that apply.

1. The domain of both functions is all real numbers.

2. The domain of f(x) is x > 5.

3. The domain of g(x) is x > 5.

4. The range of both functions is y > 5.

5. The range of f(x) is y > 0.

6. The range of g(x) is y > 0.

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Answer:

B. 8

Step-by-step explanation:

Triangle's weight is equal to the sphere's weight.

Total of 16 blocks. Since the triangle's weight is equal to the sphere's, the weight will be half of the blocks.

Divide 2 from 16.

16/2=8

The triangle's weight is equal to 8 blocks.

Hope this helps!

If not, I am sorry.

Answer:

at 5 mins Holly traveled 2 miles. At 10 mins Holly traveled 4 miles. from there on she stayed there for 10 mins then went on her way. At 30 mins she traveled 5 1/2 miles. when she hit 35 mins she hit her destination then went back home.

Step-by-step explanation:

pretty simple

Answer:

A.  [tex]\frac{21}{4} \pi\ \space\ in[/tex]

Step-by-step explanation:

The formula for arc length is:

arc length = rθ

where:

r is the radius = 7 in

θ is the central angle in radians = 135  ×  [tex]\frac{ \space\ \pi\ }{180}[/tex]  = 3/4 π

Substituting these values into the formula:
arc length = 7 in × 3/4 π

                 = [tex]\frac{21}{4} \pi\ \space\ in[/tex]

By applying the rigid transformations of reflection on y-axis, vertical compression and vertical translation, we find the transformed function based on f(x) = 2ˣ is f''(x) = (1/3) · 2⁻ˣ + 2.

Rigid transformations are transformations applied on functions such that their Euclidean distance is conserved. In this problem we must use the following rigid transformations to determine an expression based on f(x) = 2ˣ.

Reflection on y-axis

f'(x) = f(-x)     (1)

f'(x) = 2⁻ˣ

Vertical compression

f''(x) = k · f'(x)     (2)

f''(x) = (1/3) · 2⁻ˣ

Vertical translation

f'''(x) = f''(x) + c     (3)

f''(x) = (1/3) · 2⁻ˣ + 2

By applying the rigid transformations of reflection on y-axis, vertical compression and vertical translation, we find the transformed function based on f(x) = 2ˣ is f''(x) = (1/3) · 2⁻ˣ + 2.

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We'll prove first the triangles formed by diagonal are congruent then we'll prove the angles on either side of diagonal are congruent.

The steps are:

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

Answer: [tex]52.6^{\circ}[/tex]

Step-by-step explanation:

Let the angle of depression be x.

[tex]\sin x=\frac{27}{34}\\\\x=\sin^{-1} \left(\frac{27}{34} \right) \approx 52.6^{\circ}[/tex]

Answer:

6.708 miles

Step-by-step explanation:

She walked 3 miles to the right, and 6 miles up. The shortest route she could take back to her starting point is a diagonal, which we can solve for using the Pythagorean theorem, a^2 + b^2 = c^2. We're solving for c.

9 + 36 = c^2

c^2 = 45

c = 6.708

Brainliest, please :)

Answer:

14

Step-by-step explanation:

We can find the number of five-dollar bills creating a system of equations, where F represents the number of five-dollar bills and T represents the number of ten-dollar bills.  

Thus, we have 5F + 10T = 170 and F + T = 24.

We can use substitution in the second equation first to find T then finally F.

[tex]5F + 10T = 170\\F + T = 24\\\\F = 24-T\\\\5(24-T)+10T = 170\\120-5T+10T=170\\120+5T=170\\5T=50\\T=10\\\\F+10=24\\F=14[/tex]

Answer:

[tex]\huge\boxed{\sf x\leq 2}[/tex]

Step-by-step explanation:

[tex]56-10x\geq 20+8x\\\\Subtract \ 20 \ from \ both \ sides\\\\56-20-10x\geq 8x\\\\36-10x\geq 8x\\\\Add \ 10x \ to \ both \ sides\\\\36\geq 8x+10x\\\\36\geq 18x\\\\Divide \ 18 \ to \ both \ sides\\\\2 \geq x\\\\OR\\\\x\leq 2\\\\\rule[225]{225}{2}[/tex]

[tex]\bf{56-10x\geq 20+8x }[/tex]

Subtract 8x on both sides.

[tex]\bf{56-10x-8x\geq 20 }[/tex]

Combine −10x and −8x to get −18x.

[tex]\bf{56-18x\geq 20 }[/tex]

Subtract 56 from both sides.

[tex]\bf{-18x\geq 20-56 }[/tex]

Subtract 56 from 20 to get −36.

[tex]\bf{-18x\geq -36 }[/tex]

Divide both sides by −18. Since −18 is <0, the inequality direction is changed.

[tex]\bf{x\leq \dfrac{-36}{-18} }[/tex]

Divide −36 by −18 to get 2.

[tex]\bf{x\leq 2 \ \ \to \ \ \ Answer}[/tex]

We conclude that the correct option is "C".

Answer:

5

Step-by-step explanation:

substitute in 2 for x

3(2) - 1 = 5

Answer:

According to Dr. Davis' law of conversion, we are able to rewrite the equation in any order we need to so that we can solve it in multiple ways. For example, we may choose to add/subtract first, then multiply/divide. Dr. Davis discovered that when solving equations order of operations does not need to be followed.

Step-by-step explanation:

Let's suppose you're getting a new phone plan. The phone plan charges a flat fee of $5 and costs $9 a month. How can we represent this relationship?

Since $5 is a flat fee, it doesn't change based on how many months you've had the plan because it always remains $5, so this is our y-intercept.

Because the plan costs $9 a month, this represents our rate of change, or slope, showing that every month you have the plan, you multiply by $9.

So, we can show this as y=9x+5 where x is the number of months of the phone plan and y is the cost of the phone plan given x amount of months

Now, what if you wanted to know how much the phone plan would cost after 4 months?

Simple enough, we can just substitute x=4 into our equation and get y=9(4)+5=36+5=41. So, getting the phone plan for 4 months costs $41.

Let's take this the other way around. What if we wanted to figure out how many months of the phone plan are covered by $50?

We would then substitute y=50 into our equation and solve for x:

y=9x+5

50=9x+5

45=9x

5=x

This would mean $50 would cover 5 months of the phone plan.

All in all, these are real-life examples of algebra.

Considering the given graph of the function, we have that:

To find the numeric value of the function, we get a value of x(horizontal axis), and then look at the corresponding value y(vertical axis), then y = f(x).

In the graph of this function, we have that when x = 1, y = -3, hence:

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Answer:

x = 5

Step-by-step explanation:

given B is the midpoint of AC , then

AB = BC ( substitute values )

4x - 3 = 2x + 7 ( subtract 2x from both sides )

2x - 3 = 7 ( add 3 to both sides )

2x = 10 ( divide both sides by 2 )

x = 5

The probability that kids will be able to write something about both the French and Spanish classes after he is finished with the interviews is 91/100.

Given 25 kids has enrolled in atleast one foreign language class. 18 kids are in French class and 21 kids are in Spanish class.

We have to find the probability that kids will be able to write something about both the French and Spanish classes after Michael is finished with the interviews.

Let F is a set showing students knowing French,

S be a set showing students knowing Spanish ,

FS= Showing students both enrolled in one languages.

FS=F+S-(F∪S)

=18+21-25

=14

F-18-14=4

S=21-14

=7

4 outcomes lead to Michael being able to write both the languages are (F,S),(FS,S)(FS,F)(FS,FS)

Required probability is the sum of all the probabilities of (F,)(FS,S)(FS,F)(FS,FS)

P(F,S)=[tex]4C_{1}*7C_{1}/25C_{2}[/tex]

=(4*7)/300

=28/300

P(FS,S)=[tex](14C_{1}*7C_{1} )/25C_{2}[/tex]

=14*7/300

=98/300

P(FS,F)=[tex](14C_{1}*4C_{1} )/25C_{2}[/tex]

=14*4/300

=56/300

P(FS,FS)=[tex]14C_{2}/25C_{2}[/tex]

=91/300

Required probability=28/300+98/300+56/300+91/300

=(28+98+56+91)/300

=273/300

=91/100

Hence the probability that kids will be able to write both the languages is 0.91.

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The value of x is  21√2/2

The correct option is (A)

Trigonometry is a branch of mathematics that studies relationships between the sides and angles of triangles.

In triangle having angle 60.

sin 60 = P/ 7√3

√3/2= P/7√3

2P= 21

P= 21/2

Now again

cos 45= 21/2/x

1/√2 = 10.5/x

x=  21√2/2

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Using long division, 189 divided by 7 is 27

The picture below represent how to use long division.

Firstly, we divided 1 by 7, it can't go. Therefore, it zero. we put zero at the quotient space.

Multiply zero by 7 and subtract from 189.

We use 18 to divide 7. It will give us 2 remainder 4. we put the 2 at the quotient side and multiply it by 7 to get 14.

Then we subtract 14 from 18. And so on. This is done till we get 0.

Therefore, using long division, 189 divided by 7 is 27

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Answer:

Answer: x ≤ q - 17/6.

Step-by-step explanation:

Divide both sides of the inequality by the coefficient of variable