if8 kg of sugar cost Php 320.00 how much would 5kg of sugar cost?

Answer:$740

Step-by-step explanation: subtract 750 - 10 and there’s your answer look where it says “The money is taken from his bank account” that’s how you know you should subtract by the word take. Hope this helps

{11, 17, 27}

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Explanation:

Let's go through the answer choices.

Now return back to choice C. Notice how adding any two sides leads to a sum larger than the third side

The three conditions of the triangle inequality theorem are met, so a triangle is possible for choice C.

Ray is [tex]x[/tex] years old.

Ron is [tex]y[/tex] years old.

In 2 years, Ron will be two years older than now, so [tex]x+y+2[/tex] years old.

Answer:

(-∞,∞)

Step-by-step explanation:

Answer: 907.46

Step-by-step explanation:

Answer:

39

Step-by-step explanation:

all angles of a triangle equal to 180, so you just have to subtract those two known angles from 180, and you will get the answer.

Find the value of angle x .

The concept of this question belongs to angles of triangle . We know that the sum of all the three angles of triangle is 180°.

As we know that all the angles of triangle is equal to 180° . So :

Therefore , angle x is of 39° .

Answer:

the answer is

[tex] \frac{12 {c}^{8} }{ {d}^{2} } [/tex]

Step-by-step explanation:

[tex] \frac{12}{ {c}^{ - 8} {d}^{2} } [/tex]

Answer:

[tex]\frac{3\sqrt{2}+2\sqrt{3}}{6}[/tex] or [tex]\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}[/tex]

Step-by-step explanation:

[tex]\sin\frac{3\pi}{4}-\tan\frac{5\pi}{6}\\\\\sin\frac{3\pi}{4}-\frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}}\\ \\\frac{\sqrt{2}}{2}-\frac{\frac{1}{2} }{-\frac{\sqrt{3}}{2}}\\ \\\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{3}}\\ \\\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}\\ \\\frac{3\sqrt{2}}{6}+\frac{2\sqrt{3}}{6}\\ \\\frac{3\sqrt{2}+2\sqrt{3}}{6}[/tex]

Answer:

above 84°/. of the vehicles have a gas mileage greater then 24 mpg

Answer:

165

Step-by-step explanation:

area of a hexagon = 1/2 × apothem × perimeter

to find the perimeter of a hexagon we simply multiply the side length by 6 ( because there are 6 sides in a hexagon )

given side length = 11

so perimeter = 6 × 11 = 66

now to find area

remember area = 1/2 × apothem × perimeter

given apothem = 5 and perimeter = 66

so area = 1/2 × 5 × 66

1/2 × 5 = 2.5

2.5 × 66 = 165

Answer:

e) 36x² - 32y³ = (4)(9x² - 8y³)

f) 45a⁴ - 54a³ = (9a³)(5a - 6)

Step-by-step explanation:

e) 36x² - 32y³ = (4)(9x² - 8y³) becuase 4*9 = 36 and 4*8 = 32

f) 45a⁴ - 54a³ = (9a³)(5a - 6) becuase 9*5 = 45 and 9*6 = 54 and when a is multiplied by a³ you get a⁴

(the exponent are added)

PLEASE RATE!! I hope this helps!!
If you have any questions comment below

Answer:

135 degrees

Step-by-step explanation:

The image shows how i had my answer please

Answer:

The answer is D

Step-by-step explanation:

(7/x² × x²)+(1/3×x²)+(x+2/x × x²)

(7 + x/3 + x²+2x) times 3 to eliminate the 3 under x/3

so u will get 21+x+3x²+6x

Final answer = 3x²+7x+21

Answer:

sorry the image isnt clear

Answer:

Can you post it again but clearer

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Here's the solution~

Standard equation of a circle is represented as :

[tex]\qquad \sf  \dashrightarrow \:(x - h) {}^{2} + (y - k) {}^{2} = r {}^{2} [/tex]

where,

Now, let's plug in the values to find the required equation of circle ~

[tex]\qquad \sf  \dashrightarrow \:(x - 8) {}^{2} + (y - ( - 10)) {}^{2} = (4\sqrt{3} ) {}^{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \:(x - 8) {}^{2} + (y + 10) {}^{2} = 16 \times 3[/tex]

[tex]\qquad \sf  \dashrightarrow \:(x - 8) {}^{2} + (y + 10) {}^{2} = 48[/tex]

The equation of given circle with centre C ( 8 , - 10 ) and diameter ( 8√3 ) is x² + y² - 16x + 20y + 116 = 0

[tex]\quad\rule{300pt}{1pt}\quad [/tex]

The standard equation of Circle is given by :

[tex] {\pmb{\sf {\longrightarrow r^2 = (x-h)^2 +( y-k)^2}} }[/tex]

This is the standard form of the equation. Thus if we know the coordinates of center of the circle and it's radius, we can easily find its equation

Here, in this question we are given that the centre C is ( 8 , -10 ) and diameter 8√3.

[tex] \longrightarrow[/tex]h = 8

[tex] \longrightarrow[/tex] k = - 10

[tex] \longrightarrow[/tex] r = [tex] \sf{\dfrac{diameter}{2}}[/tex]

[tex] \longrightarrow[/tex]‎ r = [tex] \sf 4\sqrt{3}[/tex]

[tex] \qquad\qquad\rule{250pt}{1pt}\qquad[/tex]

On putting the values in the formula :

[tex] \sf{:\implies \qquad r^2 = (x-h)^2+ ( y-k)^2 }[/tex]

[tex] \sf{:\implies \qquad (4\sqrt{3})^2 = ( x-8)^2 +(y-(-10))^2 }[/tex]

[tex] \sf{:\implies \qquad (4)^2.(\sqrt{3})^2=(x-8)^2+(y+10)^2}[/tex]

[tex] \sf{:\implies \qquad 16\times 3 =( x^2 + 8^2 -2\times x \times 8 ) + ( y^2 + 10^2 + 2\times y \times 10 )}[/tex]

[tex] \sf{:\implies \qquad 48 = x^2 + y^2 - 16x + y^2 + 100 + 20y }[/tex]

[tex] \sf{:\implies \qquad 48 = x^2 + y^2 -16 x + 20y + 100 + 64 }[/tex]

[tex] \sf{:\implies \qquad 48 = x^2 +y^2 -16x +20y +164}[/tex]

[tex] \sf{:\implies \qquad x^2 + y^2 -16x +20y +164 - 48 = 0}[/tex]

[tex] \sf{:\implies \qquad{\boxed{ \sf x^2 + y^2 -16x + 20y +116 = 0 }}}[/tex]

‎ㅤ‎ㅤ‎ㅤ‎ㅤ~Hence, the required equation of Circle is x² + y² - 10x + 20y + 116 = 0

Answer:

D

Step-by-step explanation:

sub in the x value and if its equal to -2 its on that line

Answer:

Hope the picture will help you!!

Answer:

Step-by-step explanation:

Circumference is given by the formulas :

[tex]2\pi r[/tex] or [tex]\pi d[/tex]

Since we are given the diameter we can use the second one :

We are told to use 3.14 for π :

C = (3.14)×(16)

C = 50.24

Area of a circle is given by the formula:

[tex]\pi r^{2}[/tex]

To work out r we simply half the diamter:

16÷2 = 8

Now we put this into the area formula:

π(8)² = 64π

We are told to use 3.14 as pi so:

64π = 64×3.14 = 200.96

Answer:

Circumference = 50.24 in

Area = 200.96 in²

Step-by-step explanation:

First, let us find the circumference of the circle.

For that, let us use this formula.

C = π d

Given that,

d ⇒ diameter ⇒ 16 in

Let us find now.

C = π d

C = 3.14 × 16

C =  50.24 in

And now let us find the area of the circle.

Before that, let us find the radius.

d = 2r

16 = 2r

Divide both sides by 2.

8 in ⇒ radius

Let us use the below formula to find the area of the circle.

A = π r²

A = 3.14 × 8 × 8

A = 200.96 in²

Answer:

41, 243, 2

Step-by-step explanation:

40 + (1 + 2) - 2

=41

3[16 - (2 + 5)]^2

=243

100-50/100-3(2+3)^2

=2

Answer:

Step-by-step explanation:

General form of the linear differential equation can be written as:

[tex]\frac{dy}{dx}+P(x)y=Q(x)[/tex]

For this case, we can rewrite the equation as:

[tex]\frac{dy}{dx}+\frac{1}{2x}y=\frac{\sqrt{x}}{x}[/tex]

Here [tex]P(x) =\frac{1}{2x}; Q(x)=\frac{\sqrt{x}}{x}[/tex]

To find the solution (y(x)), we can use the integration factor method:

[tex]Fy(x)=\int Q(x)Fdx+C \rightarrow F=e^{\int P(x)dx[/tex]

Then [tex]F=e^{\int \frac{1}{2x}dx}=e^{\frac{1}{2}\ln|x|\right}=\sqrt{|x|}[/tex]

So, we can find:

[tex]y\sqrt{|x|}=\int \frac{\sqrt{x}\sqrt{|x|}}{x}dx+C[/tex]

Suppose that [tex]x\in \double R[/tex], then [tex]\sqrt{|x|}=\sqrt{x}[/tex] , and we find:

[tex]y\sqrt{x}=x+C \rightarrow y(x)=\sqrt{x}+\frac{C\sqrt{x}}{x}[/tex]

To check our solution is right or not, put your y(x) back to the ODE:

[tex]y' = \frac{1}{2\sqrt{x}}-\frac{C}{2\sqrt{x^{3}}}[/tex]

[tex]2xy'=\frac{x-C}{\sqrt{x}}[/tex]

[tex]2xy'+y=\frac{x-C}{\sqrt{x}}+\sqrt{x}+\frac{C\sqrt{x}}{x}=2\sqrt{x}[/tex]

(it means your solution is right)

Answer:

c. 3/5

Step-by-step explanation:

Using the mean concept, it is found that the correct option regarding the sample mean is given by:

b. can be smaller, larger, or equal to the population mean.

The mean of a data-set is given by the sum of all observations in the data-set divided by the number of observations.

No matter the population mean, the sample mean is calculated as stated above, having no restrictions about it's value, hence option b is correct.

More can be learned about the mean concept at https://brainly.com/question/24628525

#SPJ1

Answer:

A.  Complementary and Adjacent

B.  ∠KLM = 22°  and  ∠MLN = 68°

Step-by-step explanation:

Question A

As ∠KLN is a right angle, ∠KLN = 90°

⇒ ∠KLM + ∠MLN = 90°

When the sum of two angles is 90°, they are called complementary angles.

∠KLM and ∠MLN have a common side of ML and a common vertex (corner point) L.

When two angles have a common side and a common vertex but do not overlap in any way, the are called adjacent angles.

Question B

First find n:

∠KLM + ∠MLN = ∠KLN

⇒ 2n + (6n + 2) = 90

⇒ 2n + 6n + 2 = 90

Combine like terms:

⇒ 8n + 2 = 90

Subtract 2 from both sides:

⇒ 8n = 88

Divide both sides by 8:

⇒ n = 11

Now substitute the found value of n into the expressions for the two angles:

⇒ ∠KLM = 2(11) = 22°

⇒ ∠MLN = 6(11) + 2 = 68°

Answer:

$37.5+$3r

Step-by-step explanation:

We can assume that the number of rides is r. Since each ride is $3.00, all rides are $3r ($3xr).

We also know that there are FIVE people going to the fair, with Tory being one and each of her friends. Since the entrance fee is 7.50, we can do 7.50x5 which is $37.5.

Then, you can add it which is $37.5+$3r

Answer:

[tex]\bf C)\: -\cfrac{1}{32}[/tex]

Step-by-step explanation:

[tex]\tt \cfrac{1}{2^{-2}x^{-3}y^5}[/tex]

[tex]\tt x=2\\y=-4[/tex]

~

Substitute ''x'' with '2' & 'y' with "-4":-

[tex]\tt \cfrac{1}{2^{-2}\times \:2^{-3}\left(-4\right)^5}[/tex]

First let's solve [tex]\tt 2^{-2}\times \:\:2^{-3}\left(-4\right)^5[/tex]

[tex]\tt 2^{-2}\times \:2^{-3}=\boxed{\cfrac{1}{32} }[/tex]

[tex]\tt \cfrac{1}{32}\left(-4\right)^5[/tex]

Calculate exponents:- -4^5= -1024

[tex]\tt \cfrac{1}{32}\left(-1024\right)[/tex]

Remove parentheses, apply rule:- [tex]\tt a\left(-b\right)=-ab[/tex]

[tex]\tt-\frac{1}{32}\times \:1024[/tex]

[tex]\tt \cfrac{1}{32}\times \cfrac{1024}{1}[/tex]

Apply fraction rule:-

[tex]\tt -\cfrac{1\times \:1024}{32\times \:1}[/tex]

[tex]\tt \cfrac{1024}{32}[/tex]

Divide numbers:-

[tex]\tt -32[/tex]

Now that we're done factoring the denominator let's bring down the numerator which is ' 1 ':-

[tex]\tt -\cfrac{1}{32}\;\; \bf Answer[/tex]

☆-------☆-------☆-------☆

[tex]\dfrac 1{2^{-2}\cdot x^{-3}\cdot y^5}\\\\=\dfrac 1{2^{-2} \cdot2^{-3} \cdot (-4)^5}\\\\=-\dfrac 1{2^{-2} \cdot 2^{-3} \cdot (2^2)^5}\\\\=-\dfrac{1}{2^{-2} \cdot 2^{-3} \cdot 2^{10}}\\\\=-\dfrac{1}{2^{-2-3+10}}\\\\=-\dfrac 1{2^{5}}\\\\=-\dfrac 1{32}[/tex]

un punto en un gráfico de líneas es un valor específico que normalmente es un valor de entrada y salida (x)