please help the question is attached below

#4

Refer to the attachment

#2

Mid point:-

Step-by-step explanation:

please mark me as brainlest

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]

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:

[tex]w < 4 \frac{2}{3}[/tex]

Step-by-step explanation:

The inequality symbol for 'less than' is '<'.

Let's relate the cost and the number of each item bought together with an equation.

Total cost

= (number of rolls)(cost of each roll) +(number of bags)(cost per bag of bows)

= w(1.50) +2(1.50)

= 1.5w +3

Now, we can form the inequality.

Since the total cost is less than $10,

1.5w +3< 10

Subtract 3 from both sides:

1.5w < 10 -3

1.5w < 7

Divide both sides by 1.5:

[tex]w < 7 \div \frac{3}{2} [/tex]

[tex]w < 7 \times \frac{2}{3} [/tex]

[tex]w < \frac{14}{3} [/tex]

[tex]\textcolor{red}{w < 4 \frac{2}{3} }[/tex]

Answer:

i think the answer is d which is 50°

Answer:

Its B square pyramid

Step-by-step explanation:

Because

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:

-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:

₹ 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

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:

Let y = the amount in Adrianna's savings account

Let x = the number of weeks passed

y = 5x + 20

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

[tex]\\ \sf\Rrightarrow y=-4(x+2)^2-1[/tex]

Compare to the vertex form of parabola

So vertex should be at (h,k)=(-2,-1)

Graph attached

✿————✦————✿————✦————✿————✦————✿

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:

Step-by-step explanation:

75-52.5=x

[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]

#CarryOnLearning

#BrainlyMathKnower

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

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:

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:

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 :)

Step-by-step explanation:

a)[tex] \sf \: Factor 6x2−3x \\ \sf {6x}^{2} −3x \\ \sf \: =3x(2x−1) \\ \sf \: Answer:3x(2x−1) \\ [/tex]

b)Let's factor x²+3x−28

⇒x²+3x−28

⇒x² - 4x + 7x - 28

⇒(x² - 4x) + (7x - 28)

⇒x( x - 4) + 7(x - 4)

⇒(x -4) (x + 7)

Answer:

(x−4)(x+7)

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.

A: 1.9 Kilograms of oranges for rs 665.

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

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:

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