The length of a rectangle is 2ft longer than it’s width. If the perimeter of the rectangle is 32 ft, find it’s area

Answer:

(x+2)^2 = 9

Step-by-step explanation:

To complete the square we need to add 4 to both sides of the equation,

Why we add 4 to both sides? Because if we add it to one side only, the equation will turn into an inequation

x^2 + 4x + 4 = 5 + 4

x^2 + 4x + 4 = 9 and the factor of the expression is (x+2)*(x+2) = 9

(x+2)^2 = 9

Answer:

x = 5

Step-by-step explanation:

3^2 + 4^2 = x^2

9 + 16 = x^2

25

Now we need to find the square root of 25 :)

The answer is 5

X = 5

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The approximate diameter of the rug with an area of 40 ft² is 7.14 feet.

Area is the amount of space occupied by a two dimensional shape or object.

The area of a circle is given by:

Area = π * diameter²/4

The rug has an area of 40 square feet. Hence:

40 = π * diameter²/4

Diameter = 7.14 feet

The approximate diameter of the rug with an area of 40 ft² is 7.14 feet.

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the length of the line would be 1 inch

Answer:

37 Feet

Step-by-step explanation:

We can use Pythagoras Theorem to solve this:

[tex]a^{2} +b^{2} =c^{2}[/tex]

[tex]35^{2} +12^{2} =c^{2}[/tex]

[tex]1225+144=c^{2}[/tex]

[tex]1369=c^{2}[/tex]

[tex]\sqrt{1369} =c[/tex]

37 = c

Answer:

37 ft

Step-by-step explanation:

Use Pythagorean Theorem for right triangles to find the hypotenuse

35^2 + 12^2 = Ramp^2       ramp = 37 ft

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

Work Shown:

[tex]n = \hat{p}*(1-\hat{p})\left(\frac{z}{E}\right)^2\\\\n \approx 0.5*(1-0.5)\left(\frac{1.96}{0.03}\right)^2\\\\n \approx 1067.111\\\\n \approx \boldsymbol{1068}\\\\[/tex]

Notes:

Step-by-step explanation:

the inner angle of 2 crossing segment lines is half of the sum of the 2 arc angles.

55 = (arc angle IG + arc angle JH)/2 = (76 + JH)/2

110 = 76 + JH

arc angle JH = 34°

The required arc angle JH measures 34 degrees.

In geometry, a chord is a line segment that connects two points on the circumference of a circle. The two endpoints of the chord lie on the circle, and the chord divides the circle into two segments - the major segment and the minor segment.

Given the problem, we are given that the inner angle measures 55 degrees. Thus, we can set up an equation using the formula and solve for the unknown value.

The equation is given as follows:

55 = (arc angle IG + arc angle JH)/2

We know that the arc angle IG measures 76 degrees, so we can substitute this value into the equation:

55 = (76 + arc angle JH)/2

Multiplying both sides of the equation by 2 yields:

110 = 76 + arc angle JH

Subtracting 76 from both sides of the equation gives:

arc angle JH = 34°

Therefore, the arc angle JH measures 34 degrees.

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Answer:

Below in bold.

Step-by-step explanation:

The ratio of theire areas is  1^2 : 3^2

= 1:9.

So the area of triangle A = 28 * 9

= 252 cm^2.

Answer:

[tex]252cm {}^{2} [/tex]

step by step explanation:

[tex]since \: \: dimension \: of \: traiangle \: a \: = 3 times \: dimension \: of \: triangle \: b \\ note \: that \: area \: is \: square \: of \: dimension \: \\ therefore \: area \: of \: triangle \: a \: = 3 {}^{2} \: area \: of \: triale \: \: b \\ area \: of \: trianle \: a = 9 \times 28 = 252cm {}^{2} [/tex]

Answer:

2x = 26, x = 13

Step-by-step explanation:

Write and solve an equation twice a number is 26

x = unknown number

Equation: 2x = 26

Solve:

2x = 26

/2    /2 <== divide both sides by 2

x = 13

Check your answer:

2x = 26

2(13) = 26

26 = 26

This stament is correct

Hope this helps!

Answer:

median is 20

Step-by-step explanation:

due to both of them having it in there 1 time

Answer:

The Median Age: Yoga Class (31)      Dance Class (30)

The Range: Yoga Class (17-44)        Dance Class (20-47)

Step-by-step explanation:

Answer: 35

Step-by-step explanation:

     The average can be found by adding their ages together and dividing by the number of ages.

     Given:

[tex]\frac{19+20+20+21+x}{5}[/tex] = 23

     Multiply both sides of the equation by 5:

19 + 20 + 20 + 21 + x = 115

     Combine like terms:

80 + x = 115

     Subtract 80 from both sides of the equation:

x = 35

                    Erica is 35.

Answer:

1 3/4, 0

Step-by-step explanation:

You have to go 1 3/4 to the right and 0 up.

Answer:

84°

Step-by-step explanation:

The central angle of a circle is twice any inscribed angle subtended by the same arc.

because AB is a diameter, we also know that the triangle angle at C is 90° (all the inscribed triangles with their baseline being a diameter are right-angled).

as always, the sum of all angles in a triangle is 180°.

so, we know the triangle angle at A is

180 - 90 - 48 = 42°.

this is also the inscribed angle of the arc BC.

the central angle for arc BC (that is what is requested here) is then twice that angle : 2×42 = 84°

Answer: HELLO THERE!  Here's what I found from study dot  com            "4 divided by 5 is equal to 0.8. This decimal can also be written as a fraction. 0.8 = eight tenths or 8/10 (4/5 in its reduced form)."

Step-by-step explanation:

Hope this helps, have a good day!

Since no diagram attached there are two possible options.

Option 1

The quadrilateral is parallelogram.

In this case ∠A and ∠D sum up to 180°, therefore:

Option 2

The quadrilateral is isosceles trapezoid.

In this case ∠A is congruent with ∠D, therefore:

Answer:

115

Step-by-step explanation:

Answer:

solution 1= 40ml

solution 2= 160ml

Step-by-step explanation:

%alcohol= amount of alcohol/total solutionx100

0.52x200=104ml of alcohol present

0.2x L=0.2Lml of alcohol in solution one

0.6 x M=0.6Mml of alcohol in solution two

1st equation

0.2L+0.6M=104ml alcohol

times 10 to get whole

2L+6M=1040ml

2nd

same for this equation

10L+10M=2000ml

10L+10M=2000

2L+6M=1040

elimination method

=20L+20M=4000

-

=20L+60M=10400

-40M=-6400

M=160ml

2L+6 (160)=1040

2L=1040-960

2L=80

L=40ml

Answer:

b)

Step-by-step explanation:

Question (a)

[tex]\sf let \ (x_1,y_1)=(3,-2)\\\\\sf let \ (x_2,y_2)=(6,2)[/tex]

[tex]\sf slope \ (m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{2-(-2)}{6-3}=\dfrac43[/tex]

Point-slope formula: [tex]\sf y-y_1=m(x-x_1)[/tex]

Substituting m and point (3, -2) into the formula:

[tex]\sf y-(-2)=\dfrac43(x-3)[/tex]

[tex]\sf y+2=\dfrac43(x-3)[/tex]

Question (b)

Rewrite in standard form:

[tex]\sf y+2=\dfrac43x-4[/tex]

[tex]\sf \implies y=\dfrac43x-6[/tex]

[tex]\sf \implies 3y=4x-18[/tex]

[tex]\sf \implies -4x+3y=-18[/tex]

Answer:

The answer is 72

Step-by-step explanation:

because you take the 8 times it by the 9

9×8=72 square³

Answer:

  see attached

Step-by-step explanation:

When the signs of the x^2 and y^2 terms of a relation quadratic in both x and y are different, the relation represents a hyperbola. The graph of the hyperbola is shown in the attachment.

__

If the signs are the same, the relation represents an ellipse. If the coefficients of the squared terms are also the same, then that ellipse is a circle.

Answer:

x ≥ 4

Step-by-step explanation:

Given:

[tex]\displaystyle \large{3x-1\geq 11}[/tex]

Add both sides by 1:

[tex]\displaystyle \large{3x-1+1\geq 11+1}\\\displaystyle \large{3x\geq 12}[/tex]

Divide both sides by 3:

[tex]\displaystyle \large{\dfrac{3x}{3} \geq \dfrac{12}{3}}\\\displaystyle \large{x\geq 4}[/tex]

Hence, the solution is x ≥ 4

[tex]\textbf{Solve for u:}\\\\u(wf+v)=m+j\\\\\implies u = \dfrac{m+j}{wf+v}\\\\\textbf{Solve for f:}\\\\u(wf+v) = m+j\\\\\implies uwf+uv = m+j\\\\\implies uwf = m+j-uv\\\\\implies f=\dfrac{m+j-uv}{uw}[/tex]

Answer:

1/162

Step-by-step explanation:

first spin 1/9

land heads 50/50

second spin 1/9

so (9 x 2) x 9

= 162

    1

Answer:

1/162 or 0.62 % to the nearest hundredth.

Step-by-step explanation:

Prob(spinning a 4) = 1/9

Prob(spinning a 7) = 1/9

Prob(flipping a head) = 1/2

Required Probability = 1/9 * 1/2 * 1/9

1/162.

Answer:

You didn't attach a picture of the pentagon, but the perimiter can easily be found using one of the following formulae:

Step-by-step explanation:

P = 5s, where s is the length of a side

P = 2r × Sin(180/5), where r is the radius

The sum we want is

[tex]\displaystyle \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots[/tex]

where [tex]T_n=\frac{n(n+1)}2[/tex] is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as

[tex]\displaystyle \sum_{k=0}^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)[/tex]

For convenience, I'll use the abbreviations

[tex]S_m = \displaystyle \sum_{k=0}^\infty \frac1{(8k+m)^2}[/tex]

[tex]{S_m}' = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}[/tex]

for m ∈ {1, 2, 3, …, 7}, as well as the well-known series

[tex]\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}[/tex]

We want to find [tex]S_1-S_3-S_5+S_7[/tex].

Consider the periodic function [tex]f(x) = \left(x-\frac12\right)^2[/tex] on the interval [0, 1], which has the Fourier expansion

[tex]f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi nx)}{n^2}[/tex]

That is, since f(x) is even,

[tex]f(x) = a_0 + \displaystyle \sum_{n=1}^\infty a_n \cos(2\pi nx)[/tex]

where

[tex]a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}[/tex]

[tex]a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}[/tex]

(See attached for a plot of f(x) along with its Fourier expansion up to order n = 10.)

Expand the Fourier series to get sums resembling the [tex]S'[/tex]-s :

[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{k=0}^\infty \frac{\cos(2\pi(8k+1) x)}{(8k+1)^2} + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+2) x)}{(8k+2)^2} + \cdots \right. \\ \,\,\,\, \left. + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+7) x)}{(8k+7)^2} + \sum_{k=1}^\infty \frac{\cos(2\pi(8k) x)}{(8k)^2}\right)[/tex]

which reduces to the identity

[tex]\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'[/tex]

Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution

[tex]\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}[/tex]

It turns out that

[tex]{S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7[/tex]

so we're done, and the sum's value is [tex]\boxed{\dfrac{\pi^2}{8\sqrt2}}[/tex].

Answer:  Difference between 21 1/4 - 18 2/4 =2 3/4

Step-by-step explanation:

[tex]\text{If}~ \alpha ~\text{and}~ \beta~ \text{are the roots,}\\ \\x^2 -(\alpha + \beta)x +\alpha\beta = 0\\\\\text{For}~ x^2+3x-2=0\\\\\alpha +\beta=-\dfrac ba=-3\\ \\\alpha \beta = \dfrac ca = -2\\\\[/tex]

[tex]\text{So,~} \alpha^3+\beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 -\alpha \beta)\\ \\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -2\alpha \beta - \alpha \beta\right]\\\\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -3\alpha \beta \right]\\\\~~~~~~~~~~~~~~~~=-3\left[(-3)^2-3(-2)\right]\\\\~~~~~~~~~~~~~~~~=-3(15)\\\\~~~~~~~~~~~~~~~~=-45\\\\\text{And}~~~\alpha^3 \beta^3= (\alpha \beta )^3 = (-2)^3 = -8\\\\[/tex]

[tex]\text{Hence the equation whose roots are}~ \alpha^3 ~\text{and}~ \beta^3~ \text{is:}\\ \\ x^2-(\alpha^3 + \beta^3)x +\alpha^3 \beta^3 =0\\\\\implies x^2 -(-45)x+(-8)=0\\\\\implies x^2 +45x -8=0[/tex]

Answer:

B and B

Step-by-step explanation:

A = bh

A = 15 × 8

A = 120 in²

A = bh

A = 20 × 13

A = 260 in²

The area of the parallelogram shown are 120 in² and 260 in².

The area of a parallelogram is the product of the base and the height of the figure.

The area of a parallelogram = Base x Height

We are given that parallelogram that has a base 5 inches longer and the height 5 inches taller than a parallelogram

A = bh

A = 15 × 8

A = 120 in²

Now,

A = bh

A = 20 × 13

A = 260 in²

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Answer:

88 + 30[tex]\sqrt{7} \\[/tex]

Step-by-step explanation:

Expand to make easier:

(3[tex]\sqrt{7} \\[/tex] + 5) x (3[tex]\sqrt{7} \\[/tex] + 5)

FOIL:

9(7) + 15    [tex]\sqrt{7}[/tex] + 15[tex]\sqrt{7}[/tex] + 25

Simplify:

63 + 30[tex]\sqrt{7}[/tex] + 25

88 + 30[tex]\sqrt{7}[/tex]