!!20 POINTS!!
Find the value of X. Round to the nearest tenth

Answer:

[tex] \sqrt{41} [/tex]

Step-by-step explanation:

[tex]let \: the \: hypothenuse =\: x \\ by \: pythagoras \: theorem \\ x {}^{2} = 4 {}^{2} + 5 {}^{2} \\ x {}^{2} = 16 + 25 \\ x {}^{2} = 41 \\ sqare \: root \: bothsides \\( \sqrt{x}) {}^{2} = \sqrt{41} \\ x = \sqrt{41} \\ lenght \: of \: the \: hypothenus = \sqrt{41} [/tex]

Answer:

82 ft²

Step-by-step explanation:

Area of the patio:

5 x 9 =45

Area of the walkway:

2 x 8 =16

Area of the garden:

3 x 7 =21

Total:

45+16+21 = 82

Answer:

x=6?

Step-by-step explanation:

I think it's this answer im not sure

Answer: 24

Answer:

D. 30

Step-by-step explanation:

Notice how $2 buys 6 packets. This means that if Sandra had $1, she could buy 3 packets.

If she had $10, you must multiply 3 (packets) by 10. Sandra can buy 30 packets with $10.

Answer:

The answer is 20 (I think)

Step-by-step explanation:

The question says 6 tomato seeds at $2

So, 10*2=20

hence proved

The range of a dataset is the distance between the maximum value and the minimum value. The range is found by subtracting the minimum value from the maximum value: maximum - minimum = range.

Maximum value of this dataset = 20.4

Minimum value of this dataset = 10.1

20.4 - 10.1 = 10.3

Range = 10.3

Hope this helps!

Answer:

10.3

Step-by-step explanation:

Given set of values

Maximum value

Minimum value

Range

Answer:

23 gallons

Step-by-step explanation:

Given:

A barn is shaped like a rectangular prism with a triangular prism on top as shown.

The exterior four walls and two triangular faces need to be repainted.

To Find:

If one gallon of paint covers 232 square feet, how many gallons will it take to repaint the barn?

Solve:

[tex]12\cdot38\div2\cdot2+84\cdot20+38\cdot20\cdot2[/tex]

= [tex]5336\;square\;feet[/tex]

[tex]\frac{5336}{232}=23[/tex]

Hence 23 gallons to repaint the barn

~lenvy~

You forgot to include your slope but it should just follow the form y= mx+ b. where m is your slope and b is your y-intercept.

Recall the geometric sum,

[tex]\displaystyle \sum_{k=0}^{n-1} x^k = \frac{1-x^k}{1-x}[/tex]

It follows that

[tex]1 - x + x^2 - x^3 + \cdots + x^{2020} = \dfrac{1 + x^{2021}}{1 + x}[/tex]

So, we can rewrite the integral as

[tex]\displaystyle \int_0^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx[/tex]

Split up the integral at x = 1, and consider the latter integral,

[tex]\displaystyle \int_1^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx[/tex]

Substitute [tex]x\to\frac1x[/tex] to get

[tex]\displaystyle \int_0^1 \frac{\frac1{x^2} + 1}{\frac1{x^4} + \frac1{x^2} + 1} \frac{\ln\left(1 + \frac1{x^{2021}}\right) - \ln\left(1 + \frac1x\right)}{\ln\left(\frac1x\right)} \, \frac{dx}{x^2}[/tex]

Rewrite the logarithms to expand the integral as

[tex]\displaystyle - \int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2021}+1) - \ln(x^{2021}) - \ln(x+1) + \ln(x)}{\ln(x)} \, dx[/tex]

Grouping together terms in the numerator, we can write

[tex]\displaystyle -\int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2020}+1)-\ln(x+1)}{\ln(x)} \, dx + 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx[/tex]

and the first term here will vanish with the other integral from the earlier split. So the original integral reduces to

[tex]\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx[/tex]

Substituting [tex]x\to\frac1x[/tex] again shows this integral is the same over (0, 1) as it is over (1, ∞), and since the integrand is even, we ultimately have

[tex]\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 1010 \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 505 \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx[/tex]

We can neatly handle the remaining integral with complex residues. Consider the contour integral

[tex]\displaystyle \int_\gamma \frac{1+z^2}{1+z^2+z^4} \, dz[/tex]

where γ is a semicircle with radius R centered at the origin, such that Im(z) ≥ 0, and the diameter corresponds to the interval [-R, R]. It's easy to show the integral over the semicircular arc vanishes as R → ∞. By the residue theorem,

[tex]\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4}\, dx = 2\pi i \sum_\zeta \mathrm{Res}\left(\frac{1+z^2}{1+z^2+z^4}, z=\zeta\right)[/tex]

where [tex]\zeta[/tex] denotes the roots of [tex]1+z^2+z^4[/tex] that lie in the interior of γ; these are [tex]\zeta=\pm\frac12+\frac{i\sqrt3}2[/tex]. Compute the residues there, and we find

[tex]\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx = \frac{2\pi}{\sqrt3}[/tex]

and so the original integral's value is

[tex]505 \times \dfrac{2\pi}{\sqrt3} = \boxed{\dfrac{1010\pi}{\sqrt3}}[/tex]

Answer:

I can help you what you need help with

Step-by-step explanation:

Answer:

The farmer can predict that 55 cows in the entire population have no spots.

Step-by-step explanation:

The farmer can predict that 40 cows in the entire population have no spots, the correct option is C.

Suppose we're given that:

Favourable Cases X (in count, in sample)

Sample Size N

Level of significance =[tex]\alpha[/tex]

Then, the sample proportion of favorable cases is:

[tex]\hat{p} = \dfrac{X}{N}[/tex]

The critical value at the level of significance[tex]\alpha is Z_{1- \alpha/2}[/tex]

The corresponding confidence interval is:

[tex]CI = & \displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right)[/tex]

The given information;

The farmer took a random sample of 40 cows out of 200 total cows and found that 11 of those cows have no spots. We can use this sample data to make a prediction about the entire population using statistical inference.

Now,

Plugging in the values, we get:

CI = 0.275 ± 1.96sqrt((0.275(1-0.275))/40)

CI = 0.275 ± 0.139

CI = (0.136, 0.414)

This means that we can be 95% confident that the true proportion of cows with no spots in the entire population lies between 0.136 and 0.414.

the farmer can predict that between 13.6% and 41.4%

Therefore, by confidence interval the answer will be  farmer can predict that 40 cows in the entire population have no spots

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Answer:

87% of all the packets of biscuits Jemima displayed on the shelves.

Step-by-step explanation:

400-52=348

so, 348*25%

348*25/100

87

Answer:

the answer would be 12-19+23=18 so you need a new calculator

Step-by-step explanation:

yes

Answer:

35

Step-by-step explanation:

well, you know the angles of a triangle add up to 180 degrees, and you have the other 2 degrees right there, so all you have to do is add both of the other angles 111+34= 145 and then subtract by 180, 180-145=35 so your answer is 35

Answer:

B) B

Step-by-step explanation:

       In this case, the first partial product will be 2 * 5 which is 10. The only option that shows this is option B, so this is our answer.

       We can also check the answers as a whole,

2 * 5 = 10

80 * 5= 400

400 + 1 = 410

       Again, only option B gives us this. This is means that option B is our answer.

Answer:

B

Step-by-step explanation:

82

5

--

5 x 2 = 0 (carry 1)

5 x 8 + 1 = 1 (carry 4 to bottom)

Answer:

k

[tex]2xyy15 { { \div - \\ \sqrt[ {x. - - {?}^{?} }^{2} ]{?} }^{2} }^{2} [/tex]

Answer:

On the day the link was emailed, trailer

A

had the most views.

Two days after the link was emailed, trailer

A

had the most views.

Four days after the link was emailed, trailer

B

had the most views.

Step-by-step explanation:

The total number of views on the day the email was sent, two days later, and four days later are represented by the values of the functions when x = 0, 2, and 4.

Substitute these values into the equation:

Find these values in the table:

Apply the described rule:

So, on the day the link was emailed, trailer A had the most views.

Two days after the link was emailed, trailer A had the most views.

Four days after the link was emailed, trailer B had the most views.

Answer:

33.33 megabytes per day on average.

Step-by-step explanation:

Check the picture below.

[tex]\stackrel{\textit{\Large Areas}}{\stackrel{\textit{front and back}}{2(5\cdot 3)}~~ + ~~\stackrel{\textit{left and right}}{2(3\cdot 4)}~~ + ~~\stackrel{\textit{top and bottom}}{2(5\cdot 4)}}\implies 30+24+40\implies 94[/tex]

The area of the entire surface of the solid given is 94 units².

The area of a three dimensional object on it's outer surface is called the surface area of the object.

Given a rectangular prism.

It consists of 3 pairs of identical rectangular sides.

Area of a rectangle = length × width

Find the area of each rectangular sides to find the total area.

Total area = (5 × 3) + (5 × 3) + (5 × 4) + (5 × 4) + (4 × 3) + (4 × 3)

                 = 15 + 15 + 20 + 20 + 12 + 12

                 = 94 units²

Hence the area of the prism given is 94 units².

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Answer:

hmm i think its

1/2(x + y)

Step-by-step explanation:

because well... this would be eaiser for me if it said "of the" because "of the" is mostly used for multiply, but since it just said "the..." i can only guess. I think this may be the answer tho?

Answer:

$149 - price of the skateboard

$52 - Carly's money

$20 × 5 cars = $100

$100 + $52 = $152

$152 - $ 149 = $3

money left is $3

Answer:

529.000

Step-by-step explanation:

A decimal number is a number whose whole number and fraction part are separated by a decimal point.

529 represented as a decimal is 529.000

Also, you can add as many zeros as you want behind the decimal point.

Please pick as brainliest!

Answer:

117

Step-by-step explanation:

we should subtract 63 from 180

so 180 - 63 = 117°

Answer:

Angle CBD is 65°

Step-by-step explanation:

All of these angles add up to 180:

So, we can make this equation: (x is CBD)

46 + 21 + 48 + x = 180

115 + x = 180

-115         -115

x = 65

Hope this helps

Answer:

∠CBD = 65 degrees

Step-by-step explanation:

Please open the attached image!!

Answer:

Line LM

Step-by-step explanation:

     A ray has a point at one end, this figure does not.

     A line goes out forever in both ends, this is shown by the two arrows. Line LM describes this figure.

     Points LM are going to be two separate points, this figure does not have points.

     Plane LM will have three points to show a plane, this is a single line.

     Using these definitions and facts we know, we can figure out this is best described as Line LM.

Answer:

Step-by-step explanation:

Given that, The length of a rectangle is 3 m more than twice the width, and the area of the rectangle is 54 m²

Let's assume width of rectangle be x m and length be 3 + 2x m respectively. To find the dimensions of the rectangle we will use the formula of Perimeter of rectangle

[tex] \\ { \underline{ \boxed{ \pmb{ \sf{ \purple{Area _{(rectangle)} = Length × width}}}}}} \\ \\ [/tex]

Substituting values in above formula:

[tex] \\ \dashrightarrow \sf \: \: (3 + 2x)(x) = 54 \\[/tex]

[tex]\dashrightarrow \sf \: \: 3x + 2x^2 = 54 \\ [/tex]

[tex]\dashrightarrow \sf \: \: - 54 + 3x + 2x^2 = 0 \\ [/tex]

[tex]\dashrightarrow \sf \: \: 2x^2 + 3x - 54 = 0\\ [/tex]

[tex]\dashrightarrow \sf \: \: 2x^2 + 12x - 9x - 54 = 0\: [/tex]

[tex]\dashrightarrow \sf \: \: 2x(x + 6) -9(x + 6) = 0\\ [/tex]

[tex]\dashrightarrow \sf \: \: (2x -9)(x+6) = 0 \\ [/tex]

[tex]\sf{x=}{\sf{\dfrac{9}{2}}}{\sf{\: or\: -6}} [/tex]

[tex]\dashrightarrow \: \: { \underline{ \boxed{ \pmb{ \sf{\purple{ x = 4.5 }}}}}}[/tex]

Hence,

[tex] \\ { \underline{ \pmb{ \frak{ \therefore Length \: and \: width \: of \: the \: rectangle \: is \:12 \: m \: and \: 4.5 \: m}}}} \\ [/tex]

Answer:

Step-by-step explanation:

Given :-

To find :-

Solution:-

According to the question ,

Let the width of rectangle be x and length of rectangle be 2x+3 m

Area of rectangle :- L×B

L×B = 54m²

(2x+3) * x = 54m²

x = 4.5 m

By putting the value of x , we get

Length :- 2x+3 = 12 m

Width :- 4.5 m

Answer:

r = 8.12

Step-by-step explanation:

The volume for a cylinder is V=πr^2h therefore,

v = 2279.64

h = 11ft

r = ?

2279.64 = πr^2 * 11

[tex]\frac{2279.64}{11}= {\pi r^2} \\[/tex]

207.24 = πr^2

[tex]\frac{207.24}{\pi} =r^2\\\\66 = r^2\\\\\sqrt{66} = r\\\\\\8.12 = r[/tex]

[tex]3\frac{1}{4}[/tex]  cups of chocolate chips is 19.5 ounces.

If one cup of chocolate chips weighs about 6 ounces

we can find out the weight of [tex]3\frac{1}{4}[/tex] cups by multiplying it by the weight of one cup.

[tex]3\frac{1}{4}[/tex] cups = 3 cups + 1/4 cup

= 3 + 0.25

= 3.25 cups

Weight of 3.25 cups = 3.25 cups × 6 ounces/cup

Weight of 3.25 cups = 19.5 ounces

Therefore, [tex]3\frac{1}{4}[/tex]  cups of chocolate chips is 19.5 ounces.

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