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3 years ago

There are 40 pupils in a class. There are x more girls than boys. How many boys are there in terms of x?

Mathematics

1 answer:

lana [24]3 years ago

8 0

(40 - x) ÷ 2

this is because we first deduct the difference

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Here are the prices for cheese pizza at a certain pizzeria:

wlad13 [49]

Answer:

could u give the prices of the cheese pizza?

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3 years ago

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Suppose a family drives at an average rate of 60 mi/h on their way to visit relatives and then an average rate of 40 mi/h on the

hammer [34]

Answer:

Part A) 2 hours

Part B) d/60 hours

Part C) 48 miles/hour

Step-by-step explanation:

Part A) Let "d" be the distance in miles the family traveled to visit their relatives. How many hours did it take to drive there?

we know that

Let

x ----> the time it take on their way to visit relatives in hours

y ---> the time it take on the way back in hours

we know that

The distance of the way to visit is equal to the distance of the way back

The speed multiplied by the time is equal to the distance

60x=40y -----> equation A

y=x+1 -----> equation B

substitute equation B in equation A and solve for x

60x=40(x+1)

60x=40x+40

20x=40

x=2 hours

Find the value of y

y=2+1=3 hours

d=60x

The time it take on their way to visit relatives is 2 hours

The time it take on the way back is 3 hours

Part B) In terms of "d," how many hours did it take to drive there?

we have that

The distance is equal to the speed multiplied by the time

d=60x

Solve for x

x=d/60 hours

Part C) Write and solve an equation to determine the distance the family drove to see their relatives. What was the average rate for the entire trip?

To find the average rate for the whole trip we will need the total distance and the total time

The distance of the way to visit is

d=60x

substitute the value of x

d=60(2)=120 miles

The total distance is

120(2)=240 miles

we know that the total time is

2 + 3 =5 hours

The average rate for the complete trip is

240/5=48 miles/hour

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3 years ago

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Select an equivalent expression to:

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The answer is a.21ya

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3 years ago

Barrett is comparing the membership fees for two museums. The art museum charges a one-time fee of $8.25 plus $2.25 per month. T

sergejj [24]

Barrett will save $17.50 a year

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4 years ago

For a triangle $XYZ$, we use $[XYZ]$ to denote its area. Let $ABCD$ be a square with side length $1$. Points $E$ and $F$ lie on

nata0808 [166]

An algebraic equation enables the expression of equality between variable expressions

$\underline{The \ value \ of \ [AEF] \ is \ \dfrac{4}{9}}$

The reason the above value is correct is given as follows:

The given parameters are;

The symbol for the area of a triangle ΔXYZ = [XYZ]

The side length of the given square ABCD = 1

The location of point E = Side $\overline{BC}$ on square ABCD

The location of point F = Side $\overline{CD}$ on square ABCD

∠EAF = 45°

The area of ΔCEF, [CEF] = 1/9 (corrected by using a similar online question)

Required:

To find the value of [AEF]

Solution:

The area of a triangle = (1/2) × Base length × Height

Let x = EC, represent the base length of ΔCEF, and let y = CF represent the height of triangle ΔCEF

We get;

The area of a triangle ΔCEF, [CEF] = (1/2)·x·y = x·y/2

The area of ΔCEF, [CEF] = 1/9 (given)

∴ x·y/2 = 1/9

ΔABE:

$\overline{BE}$ = BC - EC = 1 - x

The area of ΔABE, [ABE] = (1/2)×AB ×BE

AB = 1 = The length of the side of the square

The area of ΔABE, [ABE] = (1/2)× 1 × (1 - x) = (1 - x)/2

ΔADF:

$\overline{DF}$ = CD - CF = 1 - y

The area of ΔADF, [ADF] = (1/2)×AD ×DF

AD = 1 = The length of the side of the square

The area of ΔADF, [ADF] = (1/2)× 1 × (1 - y) = (1 - y)/2

The area of ΔAEF, [AEF] = [ABCD] - [ADF] - [ABE] - [CEF]

[ABCD] = Area of the square = 1 × 1

$[AEF] = 1 - \dfrac{1 - x}{2} - \dfrac{1 - y}{2} - \dfrac{1}{19}= \dfrac{19 \cdot x + 19 \cdot y - 2}{38}$

From $\dfrac{x \cdot y}{2} = \dfrac{1}{9}$ , we have;

$x = \dfrac{2}{9 \cdot y}$ , which gives;

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)

∴ [AEF] = (1/2) × $\overline{AE}$ × $\overline{FA}$ × sin(∠EAF)

$\overline{AE}$ = √((1 - x)² + 1), $\overline{FA}$ = √((1 - y)² + 1)

[AEF] = (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)

Which by using a graphing calculator, gives;

$\dfrac{1}{2} \times \sqrt{(1 - x)^2 + 1} \times \sqrt{(1 - y)^2 + 1} \times \dfrac{\sqrt{2} }{2} = \dfrac{9 \cdot x + 9 \cdot y - 2}{18}$

Squaring both sides and plugging in $x = \dfrac{2}{9 \cdot y}$ , gives;

$\dfrac{(81 \cdot y^4-180 \cdot y^3 + 200 \cdot y^2 - 40\cdot y +4)\cdot y^2}{324\cdot y^4} = \dfrac{(81\cdot y^4-36\cdot y^3 + 40\cdot y^2 - 8\cdot y +4)\cdot y^2}{324\cdot y^2}$

Subtracting the right hand side from the equation from the left hand side gives;

$\dfrac{40\cdot y- 36\cdot y^2 + 8}{81\cdot y} = 0$

36·y² - 40·y + 8 = 0

$y = \dfrac{40 \pm \sqrt{(-40)^2-4 \times 36\times 8} }{2 \times 36} = \dfrac{5 \pm \sqrt{7} }{9}$

$[AEF] = \dfrac{9 \cdot x + 9 \cdot y - 2}{18} = \dfrac{9 \cdot y^2-2 \cdot y + 2}{18 \cdot y}$

Plugging in $y = \dfrac{5 + \sqrt{7} }{9}$ and rationalizing surds gives;

$[AEF] = \dfrac{9 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) ^2-2 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) + 2}{18 \cdot \left(\dfrac{5 + \sqrt{7} }{9}\right) } = \dfrac{\dfrac{40+8\cdot \sqrt{7} }{9} }{10+2\cdot \sqrt{7} } = \dfrac{32}{72} = \dfrac{4}{9}$

Therefore;

$\underline{[AEF]= \dfrac{4}{9}}$

Learn more about the use of algebraic equations here:

brainly.com/question/13345893

6 0

3 years ago