Someone pls

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#4

Refer to the attachment

#2

Mid point:-

Step-by-step explanation:

please mark me as brainlest

Answer:

H

Step-by-step explanation:

If you convert these fractions into decimals:
-2 1/2 = -2.5
-2.47 = -2.47
-2/5 = -0.4
5 = 5
21/4 = 5.25
You can see that the order from lest to greatest is answer H

Hope this helps :)

Answer:

  a) 96 = 3.57√h

  b) h ≈ 723.11 m

Step-by-step explanation:

The equation you want to solve is the model with the given values filled in.

  D(h) = 3.57√h . . . . model

  96 = 3.57√h . . . . . equation for seeing 96 km to the horizon

__

We solve this equation by dividing by the coefficient of the root, then squaring both sides.

  96/3.57 = √h

  h ≈ 26.891² ≈ 723.11 . . . . meters above sea level

Dustin would need to have an elevation of 723.11 meters above sea level to see 96 km to the horizon.

Answer:

170°

Step-by-step explanation:

There is a total of 180° and the angle is 10° less...

So, 180°-10° = 170°

Please 5 stars if correct!!!

I hope this helps!!!

Answer:

question 1 is 60 beads and question 2 is 16 dollars and 50 cents.

Step-by-step explanation:

for question 1 the are asking for the lowest common denomonator (60)

for question 2 you devide 60/20 and multiply by 5.50

3 x 5.50 = 16.5$

Answer:

-9/14

Step-by-step explanation:

Hey there!

Guide:

• Difference means subtract/subtraction
• Product means multiply/multiplication
• Sum means add/addition
• Quotient means divide/division

• Now that we know what “product” means… we can make the question/equation easier to solve.

3/4 × -6/7

= 3(-6) / 4(7)

= -18 / 28

= -18 ÷ 2 / 28 ÷ 2

= -9 / 14

Therefore, your answer is: -9/14

Good luck on assignment & enjoy your day!

~Amphitrite1040:)

Answer:

Let y = the amount in Adrianna's savings account

Let x = the number of weeks passed

y = 5x + 20

Answer:

Step-by-step explanation:

75-52.5=x

✿————✦————✿————✦————✿————✦————✿

So, to find the mean they added up the scores from 20 seasons and divided by 20, the number of seasons to get an average score of 10.4.

10.4 = (x1 +x2 +x3 ... +x20 )/20

Then;

10.4*(20) = (x1 +x2 +x3 ... +x20 )

208 = sum of the first 20 seasons

We want to add one more number in there, the 21st season with a score of 14. Add the 14 to the sum of the first 20 seasons and then divide by 21, the new number of seasons you're averaging.

Mean = (208+14)/21  

✿————✦————✿————✦————✿————✦————✿

Answer

= 10.6

✿————✦————✿————✦————✿————✦————✿

#carryonlearning

Answer:

Choice C - Input: 2 Output: 6

Answer:

C

Step-by-step explanation:

evaluate the output for the given input values using the rule

choice A

5 × 5 - 4 = 25 - 4 = 21 ≠ 5

choice B

3 × 5 - 4 = 15 - 4 = 11 ≠ 7

choice C

2 × 5 - 4 = 10 - 4 = 6 ← equals the output value

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

Let's think of the boat and wind as vectors:

Boat Vector --> [tex]\langle28cos36^\circ,28sin36^\circ\rangle[/tex]

Wind Vector --> [tex]\langle10cos22^\circ,10sin22^\circ\rangle[/tex]

Now, let's add the vectors:

[tex]\langle28cos36^\circ+10cos22^\circ,28sin36^\circ+10sin22^\circ\rangle[/tex]

Find the magnitude (the true velocity):

[tex]\sqrt{(28cos36^\circ+10cos22^\circ)^2+(28sin36^\circ+10sin22^\circ)}\approx37.78\approx38[/tex]

Find the direction (angle):

[tex]\theta=tan^{-1}(\frac{28sin36^\circ+10sin22^\circ}{28cos36^\circ+10cos22^\circ})\approx32.32^\circ\approx32^\circ[/tex]

Thus, D is the best answer

Problem 2

Recall that the angle between two vectors is [tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})[/tex] where [tex]u\cdot v[/tex] is the dot product of the vectors and [tex]||u||*||v||[/tex] is the product of each vector's magnitude:

[tex]\theta=cos^{-1}(\frac{u\cdot v}{||u||*||v||})\\\\\theta=cos^{-1}(\frac{\langle-82,47\rangle\cdot\langle92,80\rangle}{\sqrt{(-82)^2+(47)^2}*\sqrt{(92)^2+(80)^2}})\\\\\theta=cos^{-1}(\frac{(-82)(92)+(47)(80)}{\sqrt{6724+2209}*\sqrt{8464+6400}})\\\\\theta=cos^{-1}(\frac{(-7544)+3760}{\sqrt{8933}*\sqrt{14864}})\\\\\theta=cos^{-1}(\frac{-3784}{\sqrt{132780112}})\\\\\theta\approx109.17^\circ\approx109^\circ[/tex]

Therefore, C is the best answer

Answer:

ONE “1”

Step by step:

Taking into account the definition of a system of equations, this nonlinear system of equations has one solution.

A system of equations is a set of two or more equations that share two or more unknowns. The solutions of a system of equations are all the values that are valid for all equations.

That is, to find the solution to a system of equations, you must find a value (or range of values) that satisfies all the equations in the system.

Graphically, the points where the graphs of the equations intersect will be the solutions to the system.

A system of equations with a linear equation and a quadratic equation can have two, one, or no solutions. To find its quantity, we must look for the intersection between both graphs:

If the graphs of the equations do not intersect, then there are no solutions for both equations. If the parabola and the line touch at a single point, then there is a solution for both equations. If the line intersects the parabola in two places, then there are two solutions to both equations.

In this case, you can see that the parabola and the line touch at a single point. Then there is a solution for both equations.

In summary, this nonlinear system of equations has one solution.

Answer:

[tex](9x-1)^2[/tex]

Step by step explanation:

[tex]81x^2-18x+1\\\\=81x^2 - 9x-9x+1\\\\=9x(9x-1)-(9x-1)\\\\=(9x-1)(9x-1)\\\\=(9x-1)^2[/tex]

Answer:

14cm

Step-by-step explanation:

Algebra is the way!

But seriously, let's model this algebraically.

We know the that the area of a triangle is represented by:

[tex]A = \frac{1}{2} base*height[/tex]

we'll use b and h to represent base and height

[tex]A = \frac{1}{2} b*h[/tex]

Okay, so you are given the area of the triangle already. It is [tex]35 cm^2[/tex]. SO you have:

[tex]35 = \frac{1}{2} bh[/tex]

Now here's the tricky bit of the question. You have 2 unknown variables, oh no! How will you solve? The key to this is simple, in a case where you have 2 or more unknown variables, ALWAYS think to yourself, "can I represent this variable in terms of another"?

In this case you can. They tell you what the height of the triangle is, in terms of the length of the base. They tell you, height is 4 cm more than 2 times the base.

So replace h in this equation with 2b + 4.

[tex]35 = \frac{1}{2} (b)(2b+4)\\\\35 = \frac{1}{2}(2b^2 + 4b)\\ 70 = 2b^2 + 4b\\0 = 2b^2 + 4b - 70\\0 = (2)(b^2 + 2b - 35)\\0 = (b +7)(b - 5)\\\\b = 5\\b\ \neq -7[/tex]

Okay, so you know the base of the triangle is 5 cm. The question asks for height, so what you're going to do is refer back to what h is.

We found that [tex]h = 2b + 4[/tex]

sub in 5 into that:

[tex]h = 2(5) + 4 = 14[/tex]

Therefore the height of the triangle is 14cm.

Answer:

Step-by-step explanation:

Givens

Area = 1/2 B* H

B = B

H = 2*B + 4

Area = 35

Solution

35 = 1/2 * B * (2B + 4)                          Multiply both sides by 2

35*2 = 2*(1/2) * B * (2B + 4)                 Combine

70 = B(2B + 4)                                      Remove the Brackets.

70 = 2B^2 + 4B                                    Subtract 70 from both sides.

2B^2 + 4B - 70 = 0                               Divide everything by 2

B^2 + 2B - 35 = 0                                 Factor

(B + 7)(B - 5)

B+7 = 0 will give a minus number. Geometry does not allow minus numbers.

B - 5  = 0

B = 5

But what you want is the height.

h = 2b + 4

h = 2*5 + 4

Answer: h = 14

The tree grew about 3.83 inches between the first and third week.

An equation is an expression that shows the relationship between two or more numbers and variables.

On the first week, the tree was 27 1/2 inches (27.5) tall. On the second week, it was 29 1/4 inches (29.25) tall. On the third week, it was 31 1/3 inches (31.33) tall.

The length grown = 31.33 - 17.5 = 3.83 inches

The tree grew about 3.83 inches between the first and third week.

Find out more on equation at: https://brainly.com/question/2972832

Answer:

₹ 81.75

Step-by-step explanation:

Total =  ₹ 120.

He bought one book for ₹ 25.75 and a pen for ₹ 12.50.

    ₹ 25.75 + ₹ 12.50 =  ₹ 38.25

How much money does he have now?​

     ₹ 120 -  ₹ 38.25 =  ₹ 81.75

₹81.75

you just minus 120 from 25.75 and 12.50

Using the Pythagorean Theorem, it is found that Barrie is 12 miles from Luke's house, in a straight line.

The Pythagorean Theorem relates the length of the legs [tex]l_1[/tex] and [tex]l_2[/tex] of a right triangle with the length of the hypotenuse h, according to the following equation:

[tex]h^2 = l_1^2 + l_2^2[/tex]

This problem can be modeled by a right triangle, with [tex]l_1 = 16, l_2 = d, h = 20[/tex], hence:

[tex]h^2 = l_1^2 + l_2^2[/tex]

[tex]16^2 + d^2 = 20^2[/tex]

[tex]d^2 = 144[/tex]

[tex]d = 12[/tex]

Barrie is 12 miles from Luke's house, in a straight line.

More can be learned about the Pythagorean Theorem at https://brainly.com/question/654982

Answer:

7.1 x 10³

Step-by-step explanation:

Move your decimal so that it follows your first whole number greater than zero. The number of places you moved the decimal is the same as the exponent. Since the number 7.1 needs to INCREASE to be 7,100, then the exponent is positive.

Answer:

7.1 x 10[tex]^{3}[/tex] (note the 3 is an exponent)

Step-by-step explanation:

1. Count how many places there are to the right of 7

2. There are 3

3. Apply the formula of scientific notation a x 10[tex]^{b}[/tex]

4. 7.1 x 10[tex]^{3}[/tex]  (remove all digits after 7.1  

Hope this helps :)

Answer:

7x - 3, x being the number

Step-by-step explanation:

number = x

7x - 3

Answer:

Three less than 7 times a number is 39.

Step-by-step explanation:

Answer:

i think the answer is d which is 50°

Answer:

2

Step-by-step explanation:

To find the interquartile range, subtract the lower quartile from the upper quartile.

The upper quartile is 7.

The lower quartile is 5.

Subtract.7–5=2

The interquartile range for Mike's teammates' shoe sizes is 2 sizes.

The interquartile range for Mike's teammates' shoe sizes is R = 2

The distance between the upper and lower quartiles is known as the interquartile range. Half of the interquartile range corresponds to the semi-interquartile range. Finding the values of the quartiles in a small data set is straightforward.

The spread of the data, or statistical dispersion, is measured by the interquartile range. The middle 50%, fourth spread, or H-spread are further names for the IQR. It is described as the spread between the data's 75th and 25th percentiles.

Given data ,

Let the interquartile range be represented as R

Now , the equation will be

Let the shoe sizes be represented as set A

A = { 2 , 4 , 6 , 8 , 10 }

Now , the upper range of the quartile is = 7

The lower range of the quartile is = 5

And , the interquartile range R = upper quartile - lower quartile

On simplifying , we get

The interquartile range R = 7 - 5 = 2

Hence , the interquartile range R = 2

To learn more about interquartile range click :

https://brainly.com/question/15686744

#SPJ3

Answer: C

Step-by-step explanation:

Just look at a z-score table and multiply by 100.

-> (0.308538)(100) is about 30.85%

[tex] \Large \bold{SOLUTION\ 1:} [/tex]

[tex] \small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array} [/tex]

[tex] \small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array} [/tex]

[tex] \small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

[tex] \Large \bold{SOLUTION\ 2:} [/tex]

[tex] \small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array} [/tex]

[tex] \small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

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#BrainlyMathKnower

A: 1.9 Kilograms of oranges for rs 665.

B: 12 liters of petrol will make a car travel 120 km.

Step-by-step explanation:

2× (4x-5y) = 21

3x+10y= -53

-------------------------

8x-10y = 42

3x+10y=-53

------------------------

11x= -11

{x=-1}

-----------------

3x+10y=-53

x=-1

3(-1)+10y=-53

-3+10y=-53

10y= -50

{y=5}

Answer:

[tex] \frac{ - 1}{6} [/tex]

Step-by-step explanation:

[tex]slope \: = \frac{y2 - y2}{x2 - x1} \\ slope = \frac{3 - 4}{5 - - 1} \\ sope = \frac{ - 1}{ \: \: \: 6} [/tex]