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3 years ago

during the soccer season rick scored on 40% of the shots he took. if scored 12 goals, how many shots did he take

Mathematics

1 answer:

UkoKoshka [18]3 years ago

8 0

The answer is 30 because 40% of 30 is 12

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Out of the 2000 students at Smith Middle School, 250 have green eyes, which is 12.5%. Mrs. Craig surveys her math class and 8 ou

vivado [14]

B)
The sample that Mrs. Craig took was too small.

I'm assuming the question is asking why they were not the same percentage?

6 0

4 years ago

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2xB+4= -8 what number does B stand for in the question

IgorC [24]

2 x B + 4 = -8
-4 -4
2 x B = -12
divide by 2 on both sides
B = -6

4 0

4 years ago

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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

What are the approximate solutions of 4x2 + 3 = −12x to the nearest hundredth?

Gwar [14]

First
$4x^{2} + 12x + 3 = 0$
determine if this equation were factorable, it's not for this one
so I use quadratic formula, try to search the formula online

$x1 = \frac{-12 + \sqrt{(12)^{2} - 4*4*3} } {2*4} = \frac{-12 + 4\sqrt{6} }{8} = \frac{4(-3 + \sqrt{6}) }{8} = \frac{-3 + \sqrt{6}}{2} = - 0.28$
$x2 = \frac{-12 - \sqrt{(12)^{2} - 4*4*3} } {2*4} = \frac{-12 - 4\sqrt{6} }{8} = \frac{4(-3 - \sqrt{6}) }{8} = \frac{-3 - \sqrt{6}}{2} = - 2.72$

x ≈ −2.72 and x ≈ −0.28 is the solution

3 0

4 years ago

Christina's purchasing a new TV. She was approved to finance the TV with zero interest. If Christina gives a one-time payment of

Allisa [31]

Given:

One time payment, p = $300

Payment per month, q = $65

Number of months paid, n = 5

The objectiv is to find the amount she paid in 5 months.

Let x be the amount she paid in 5 months. Then the the formula is,

$x=p+nq$

Let's substitute the values.

$\begin{gathered} x=300+5(65) \\ x=300+325 \\ x=625 \end{gathered}$

Hence, total amount paid in 5 months is $625.

3 0

1 year ago