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3 years ago

What is the measure of

Mathematics

1 answer:

Annette [7]3 years ago

8 0

Answer:

gvb

Step-by-step explanation:

your question is wrong

ask right question first

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4x + 8y = 20 -4x + 2y = -30 solve using elimination

Ahat [919]

(7,-1)
comment if u want me to explain the answer

8 0

2 years ago

A teenager who is 5 feet tall throws an object into the air. The quadratic function LaTeX: f\left(x\right)=-16x^2+64x+5f ( x ) =

tia_tia [17]

Answer:

At approximately x = 0.08 and x = 3.92.

Step-by-step explanation:

The height of the ball is modeled by the function:

$f(x)=-16x^2+64x+5$

Where f(x) is the height after x seconds.

We want to determine the time(s) when the ball is 10 feet in the air.

Therefore, we will set the function equal to 10 and solve for x:

$10=-16x^2+64x+5$

Subtracting 10 from both sides:

$-16x^2+64x-5=0$

For simplicity, divide both sides by -1:

$16x^2-64x+5=0$

We will use the quadratic formula. In this case a = 16, b = -64, and c = 5. Therefore:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute:

$\displaystyle x=\frac{-(-64)\pm\sqrt{(-64)^2-4(16)(5)}}{2(16)}$

Evaluate:

$\displaystyle x=\frac{64\pm\sqrt{3776}}{32}$

Simplify the square root:

$\sqrt{3776}=\sqrt{64\cdot 59}=8\sqrt{59}$

Therefore:

$\displaystyle x=\frac{64\pm8\sqrt{59}}{32}$

Simplify:

$\displaystyle x=\frac{8\pm\sqrt{59}}{4}$

Approximate:

$\displaystyle x=\frac{8+\sqrt{59}}{4}\approx 3.92\text{ and } x=\frac{8-\sqrt{59}}{4}\approx0.08$

Therefore, the ball will reach a height of 10 feet at approximately x = 0.08 and x = 3.92.

7 0

2 years ago

Which function is f(x) equal to f^-1(x)?

Morgarella [4.7K]

Answer:

c. f(x) x+1/x-1

Step-by-step explanation:

To answer this question, we need to check each answer one by one until we find the right one.

y = (x+6)/(x-6)

switch x and y

x = (y+6)/(y-6)

solve for y

x(y-6) = y+6

xy - 6x = y+6

y(x-1) = 6x+6

y = (6x+6) /(x-1) = 6(x+1)/(x-1)

f^-1(x) = 6(x+1)/(x-1)

y = (x+2)/(x-2)

switch x and y

x = (y+2)/(y-2)

solve for y

x(y-2) = y+2

xy -2x = y+2

y(x-1) = 2x+2

y = (2x+2)/(x-1)

f^-1(x) = 2(x+1)/(x-1)

y = (x+1)/(x-1) ------ correct one

switch x and y

x = (y+1)/(y-1)

solve for y

x(y-1) = y+1

xy - x = y+1

y(x-1) = x+1

y = (x+1)/(x-1)

f^-1(x) = (x+1)/(x-1)

f(x) = f^-1(x)

6 0

3 years ago

Read 2 more answers

one rectangle has a length 10 cm and width 5 cm the second rectangle has length 12 cm and width 4 cm are the two rectangles simi

tatuchka [14]

Answer:

No, they aren't similar.

Step-by-step explanation:

Let's take the proportionality of their sides to check whether they are similar or not:

$\frac{10}{5} = \frac{12}{4}$

=> 2 ≠ 3

So, they are not similar.

8 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}$

$\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago