The price of a cup of coffee was 2.40 yesterday. Today, the price rose to
2.75. Find the percentage increase. Round your answer to the nearest tenth of a percent.

From the slope, it is possible to classify the lines as parallel.

A line can be represented by a linear function. The standard form for the linear equation is: ax+b , for example, y=2x+7. Where:

a= the slope

b=constant term that represents the y-intercept.

The lines are classified as parallel when they have the same slope. You will have perpendicular lines when one line is the negative reciprocal of the second line. If the lines intersect, none of the previous conditions presented are met.

The given equations are:

6x-2y=-2 (1)

y=3x+12 (2)

First, you should write the equation 1 in the standard form.

-2y=-6x-2

-y=-3x-1

y=3x+1

Therefore, the slope for equation 1 is m=3.

After that, you should indicate the slope for equation 2. Thus, m also is 3.

As, both equations present the same slope (m=3). The lines are parallel.

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Answer: The Product(?) of their slopes is 3, so the lines are parallel

Step-by-step explanation:

summary of what the other person said

Answer:

1/3 and 1/4

multiplying 1/3 by 1/4 and 1/4 by 1/3 we get

1/3 * 1/4

1 *1 /3*4

1/12

1/4 * 1/3

1*1 / 4*3

1/12

now, both have same denominator.

This can be represented by [tex]\boxed{10(x+y)}[/tex]

Answer:

16 : 19

Step-by-step explanation:

Given weights of the two castings : 32 lbs and 38 lbs respectively.

Now,

Ratio of the weights of the two castings ;

= 32 : 38

= 16 : 19

32:38

16:19

Answer:

this is my answer

Step-by-step explanation:

volume= l×w×h

1.1/3×1/2×2/3

=1/9

2. 3/4×1/8×13/16

=39/512

3. 1/12×7/8×1/4

=7/384

4. 3/2×1/7×7/8

=3/16

5. 1/5×9/10×2/3

=3/25

6.6/7×7/8×1/2

=3/8

1) [tex]V=lwh=5(x)(8)=\boxed{40x}[/tex]

2) [tex]V=\pi r^{2}h=(\pi)(3^{2})(8)=\boxed{72}\pi[/tex]

3) [tex]40x=72\pi\\\\x=\frac{72\pi}{40} \approx \boxed{5.7}[/tex]

Answer:

See below

Step-by-step explanation:

x2 + 4x + 12 = 0    Complete the square ( take 1/2 of the 'x' coefficient, then square it   and add it to both sides of the equation

x^2 + 4x + 2^2 + 12 = 2^2     Subtract 12 from both sides

x^2 + 4x + 4 = -8

(x+2)^2 = - 8       sqrt both sides

(x+2) =± sqrt (-8)     simplify just a bit

x+2 = ±(2i sqrt2)      subtract 2 from both sides

x =  -2 ± 2i sqrt 2       <===== I do not see that answer listed

The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

For a sector that has a central angle of θ, and a radius of r;

The sector area, and the arc length are:

[tex]A = \frac{\theta}{360} * \pi r^2[/tex] --- sector area

[tex]L = \frac{\theta}{360} * 2\pi r[/tex] ---- arc length

Here, the given parameters are:

Central angle, θ = 160

Radius, r = 5 inches

The sector area is

[tex]A = \frac{\theta}{360} * \pi r^2[/tex]

So, we have:

[tex]A = \frac{160}{360} * \frac{22}{7} * 5^2[/tex]

Evaluate

A = 34.92

The arc length is:

[tex]L = \frac{\theta}{360} * 2\pi r[/tex]

So, we have:

[tex]L = \frac{160}{360} * 2 * \frac{22}{7} * 5[/tex]

L = 13.97

Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

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Answer:

A set is a well-defined collection of distinct mathematical objects.

Answer:

[tex]9600cm^{2}[/tex]

Step-by-step explanation:

Get the area of the triangle and the rectangle, then subtract the triangle's area for the rectangle's

Answer: y = -x + 7

Step-by-step explanation:

The slope is [tex]\frac{5-6}{2-1}=-1[/tex], so we know the equation is of the form [tex]y=-x+b[/tex].

Substituting in the coordinates (1,6) to find b,

[tex]6=-1+b\\\\b=7[/tex]

Thus, the equation is y = -x + 7

The equation is:

(x-5)^2+(y-(-7))^2=36

I hope it helps!

Answer:  [tex]\boldsymbol{(x - 5)^2 + (y + 7)^2 = 36}[/tex]

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

Work Shown:

[tex](h,k) = (5,-7) = \text{center}\\\\r = 6 = \text{radius}\\\\(x - h)^2 + (y - k)^2 = r^2\\\\(x - 5)^2 + (y - (-7))^2 = 6^2\\\\\boldsymbol{(x - 5)^2 + (y + 7)^2 = 36}\\\\[/tex]

Answer:

in exploratory factor analysis all the measures variables are related to every legend variable but in confirmatory factor analyse CFA research 10 specifically the number of factors required in the data in which measures variable is related with talent variable

The value could the length of the third side will be 11,12,13,15. Options B, D, E, and F are the correct answers.

A triangle is a three-sided polygon. It is one of the most fundamental geometric forms.

According to the triangle Inequality theorem, any two triangle sides must add up to more than the third side's length.

You may get the potential length of a triangle by multiplying the sum of two sides by the difference between two sides.

Therefore,

Let x is the possible length of the third side of a triangle.

14-4<x<15+4

10<x<19

The values of the third side is lies between the 10 and 19.The value could the length of the third side will be 11,12,13,15

Hence, options B, D, E, and F are the correct answers.

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Answer:

Your answer would be A.

Step-by-step explanation:

okay so, with a best fit line, it needs to match up with your points. unless your points go up evenly. its not going to connect to all your points.

Answer:

5.29

Step-by-step explanation:

8

[tex] \sqrt{8 { = }^{2} { - 6 {?}^{2} }^{2} } [/tex]

Answer: Missing side length (b) = 5.3 units

Step-by-step explanation:

Hii, I'd love to help you out! (:

We are provided with two sides, which are:

Luckily, there's a formula to find the third side!

The Pythagorean Theorem ([tex]\pmb{a^2+b^2=c^2}[/tex]) is the formula that we will use to find the missing side. In this formula "a" and "b" denote the "legs" of the triangle (the two shorter sides) and "c" denotes the hypotenuse (the longest side which is  opposite the right angle (the 90 degree angle)

NB: a and b are interchangeable, because a+b is the exact same thing as b+a; c is always the hypotenuse.

Now it's time for us to stick in the values.

[tex]\bigstar\underbrace{\boxed{\sf a=6;\;c=8}}[/tex]

This is what we obtain upon sticking in the values:

[tex]\tt 6^2+b^2=8^2[/tex]

Now we can solve for b.

[tex]\tt 36+b^2=64}\\b^2=64-36\\b^2=28\\b\approx5.3[/tex]

Thus, the value of "b" is 5.3. (Rounded to the nearest tenth, or one decimal place - one digit after the decimal point)

_____________

Hope I helped! Best wishes.

Reach far. Aim high. Dream big.

_____________

[tex]\underbrace[/tex]

Answer:

24.012

Step-by-step explanation:

y=kx^2

54=81k

k=54/81

k=0.667

y=kx^2

y=0.667×36

y=24.012

its false ik this because im one of the boyzz

The mean and the standard deviation are 16.15 and 12.03, respectively

The table of values is given as:

Work hours           Frequency

0 - <10                        45

10 - <20                      33

20 - <30                     20

30 - <40                       7

40 - <50                      8

Start by rewriting the table using the midpoint of each class

x      f

5     45

15    33

25    20

35    7

45    8

The mean is then calculated as:

[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]

This gives

[tex]\bar x = \frac{5 * 45 + 15 * 33 + 25 * 20 35 * 7 + 45 * 8}{45 + 33 + 20 + 7 + 8}[/tex]

Evaluate the sum and the products

[tex]\bar x = \frac{1825}{113}[/tex]

Divide

[tex]\bar x = 16.15[/tex]

The standard deviation is:

[tex]\sigma =\sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}[/tex]

This gives

[tex]\sigma= \sqrt{\frac{45 * (5 - 16.15)^2 + 33 * (15 - 16.15)^2 + 20 * (25 - 16.15)^2 + 7 * (35 - 16.15)^2 + 8 * (45 - 16.15)^2 }{45 + 33 + 20 + 7 + 8}}[/tex]

Evaluate the sum and the products

[tex]\sigma= \sqrt{\frac{16350.4425}{113}}[/tex]

Divide

[tex]\sigma= \sqrt{144.694181416}[/tex]

Take the square root

[tex]\sigma= 12.03[/tex]

Hence, the mean and the standard deviation are 16.15 and 12.03, respectively

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The approximate appreciation or depreciation we might see in the MXN/JPY cross exchange rate is 3%.

The term depreciation refers to an accounting method used to allocate the cost of a tangible or physical asset over its useful life. Depreciation represents how much of an asset's value has been used.

The approximate appreciation or depreciation we might see in the MXN/JPY cross exchange rate can be stated using the following 3 steps.

Step 1. State the initial exchange rates of the currency pairs.

Let first assume the initial exchange rates are as follows:

USD1 = JPY1

USD1 = MXN1

Therefore, we have the initial cross rate as follows:

MXN1 = USD1 = JPY1

MXN1 = JPY1

Step 2. Determine the new exchange rates

The new exchange rates can be determined as follows:

When the USD depreciates 2% versus the JPY, this implies that

USD1 X (100% + 2%) = USD1.02

has to be exchanged for JPY1.

Therefore, we now have:

USD1.02 = JPY1, or

USD1 = JPY1/1.02

USD1 = JPY0.98

Also, when The USD appreciates 1% versus the MXN, this implies that USD1  X  (100% - 1%) = USD0.99

has to be exchanged for MXN1.

Therefore, we now have:

USD0.99 = MXN1, or

USD1 = MXN1/0.99

USD1 = MXN1.01

Therefore, we have the new cross rate as follows:

MXN1.01 = USD1 = JPY0.98

MXN1.01 = JPY0.98

MXN1.01 / 1.01 = JPY0.98/1.01

MXN1 = JPY0.97, or

MXN1/0.97 = JPY0.97/0.97

MXN1.03 = JPY1

Therefore, the new exchange rates are as follows:

USD1.02 = JPY1

USD0.99 = MXN1

MXN1.03 = JPY1

c. Determination of appreciation or depreciation we might see in the MXN/JPY

Percentage of depreciation of MXN against JPY = ((Initial MXN/JPY - New MXN/JPY) / Initial MXN/YPY)  X 100 = ((1.03 - 1) / 1) * 100 = 3%

Since the percentage of depreciation of MXN against JPY is 3%, this also implies that the percentage of appreciation of JPY against MXN is 3%.

Thus, the approximate appreciation or depreciation we might see in the MXN/JPY cross exchange rate is 3%.

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The equation fourth represents the ellipse that has vertices at (1, 0) and (11, 0), and an eccentricity of 3/5 option fourth is correct.

An ellipse is a locus of a point that moves in a plane such that the sum of its distances from the two points called focus adds up to a constant. It is taken from the cone by cutting it at an angle.

We have:

The vertices of the ellipse = (1, 0) and (11, 0)

Eccentricity of ellipse = 3/5

We have equations:

[tex]\rm \dfrac{\left(x-6\right)^{2}}{100}+\dfrac{y^{2}}{64}=1[/tex]

[tex]\rm \dfrac{\left(x+6\right)^{2}}{64}+\dfrac{y^{2}}{100}=1[/tex]
[tex]\rm \dfrac{\left(x+6\right)^{2}}{16}+\dfrac{y^{2}}{25}=1[/tex]

[tex]\rm \dfrac{\left(x-6\right)^{2}}{25}+\dfrac{y^{2}}{16}=1[/tex]

From the fourth equation:

The vertices of the ellipse:

(1, 0) and (11, 0)

The eccentricity:

c²  = a² - b²

c² = 25-16

c = 3

e = c/a = 3/5

Thus, the equation fourth represents the ellipse that has vertices at (1, 0) and (11, 0), and an eccentricity of 3/5 option fourth is correct.

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If the angle 1 and angle 2 are vertical angles, hence they must be equal to each other, therefore m∠1 = m∠2 = 72º

Angles angle are angles that are equal to equal to each other in a line geometry.

Given the following parameters

m∠1 = 72º,

If the angle 1 and angle 2 are vertical angles, hence they must be equal to each other, therefore m∠1 = m∠2 = 72º

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The percent increase of women students is 21.13%.

Given that, an area vocational school had an enrollment of 325 men and 123 women.

Percentage is defined as a given part or amount in every hundred. It is a fraction with 100 as the denominator and is represented by the symbol "%".

In 2006, there were 149 women.

Percentage increase = (New value - original value)/Original value × 100

= (149-123)/123 × 100

= 26/123 × 100

= 21.13%

Therefore, the percent increase of women students is 21.13%.

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Answer:

3x²

Step-by-step explanation:

0.482 to 1sf is 0.5

61.2x² to one sf is 60x²

√98.01 to 1sf is √100

√100 = 10

0.5 x 60x² = 30x²

30x² / 10 = 3x²

Answer:

x=80.8

Step-by-step explanation:

follow BODMAS

2015-(2015÷5(x-5)÷5)=2014

2015-(2015÷5x-25÷5)=2014

2015-(405/x-5)=2014

2015-(405-5x)=2014

2015-405+5x=2014

1610+5x=2014

5x=2014-1610

(5x=404)1/5

x=80.8

Let

[tex]I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]tex

Integrate by parts:

[tex]\displaystyle \int u \, dv = uv - \int v \, du[/tex]

with

[tex]u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)[/tex]

Then

[tex]\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C[/tex]

Integrate by parts again, this time with

[tex]u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)[/tex]

so that

[tex]\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}[/tex]

Answer:

x = -3

Explanation:

[tex]\rightarrow \sf \sqrt[3]{\sf 3x + 1} = -2[/tex]

[tex]\rightarrow \sf \sf 3x + 1 = (-2)^3[/tex]

[tex]\rightarrow \sf \sf 3x + 1 = -8[/tex]

[tex]\rightarrow \sf \sf 3x = -8-1[/tex]

[tex]\rightarrow \sf \sf 3x = -9[/tex]

[tex]\rightarrow \sf \sf x = -3[/tex]

Answer:

[tex]\sqrt[3]{3x+1} =-2\\(\sqrt[3]{3x+1})^{3} =(-2)^{3} \\3x+1=-8\\3x=-9\\x=-3[/tex]

Hope it helps!

Answer:

x = 5/2

Step-by-step explanation:

Given that,

[tex] \cfrac{x}{3} = \cfrac{5}{6} [/tex]

Solution:

Multiplying using criss-cross method;

Divide both sides by 6;

Done!

Hence x = 5/2.

Answer:

x = 2.5

Step-by-step explanation:

x = 3*(5/6)