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2 years ago

Which statement about the function below is TRUE?

Mathematics

2 answers:

fenix001 [56]2 years ago

3 0

The statement b is true. A function is increasing when the slope of the tangent is positive. Between 2 and 3, the function has a positive slope so it is increasing.

gavmur [86]2 years ago

3 0

Its increasing so it is true

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Alex gets paid $2.25 a tree for tree planting up north. In one week, he planted 437 trees. What is his wage?

Amiraneli [1.4K]

Answer:

Hey Beebzie

Dang he part of 20m trees? And is Alex a he or she, we'll never know, so I'll just refer to Alex as "it" to not gender appropriate anyone.

So it gets paid 2.25 for one tree. It planted 437 tress.

We take 2.25 x 437 to know the amount of money earned for all trees it planted.

The answer is $983.25, or A. I don't want to gender appropriate the letter "A" either because it could be option 1 or a different letter so I'll just say the answer again to not be sexist, $983.25

6 0

2 years ago

(6 to the power of 3 times 3 to the power of 3) to the power of 2

frosja888 [35]

Answer:

True

Step-by-step explanation:

(6³ * 3³)² = 18⁶

Multiply inside the parenthesis.

(18³)² = 18⁶

When you put powers to powers, they multiply, eg:

(x³)⁴ = x³°⁴ = x¹²

18⁶ = 18⁶

True.

4 0

2 years ago

Question:

Andrew [12]

Answer:

(0, -9)

Step-by-step explanation:

The y intercept is the y value when x =0

(0, -9)

4 0

2 years ago

Select the correct answer from each drop-down menu.

serious [3.7K]

Answer:

Center: (4,8)

Radius: 2.5

Equation: $(x-4)^2+(y-8)^2=6.25$

Step-by-step explanation:

It was given that; the endpoints of the longest chord on a circle are (4, 5.5) and (4, 10.5).

Note that the longest chord is the diameter;

The midpoint of the ends of the diameter gives us the center;

Use the midpoint formula;

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

The center is at; $(\frac{4+4)}{2} ,\frac{5.5+10.5}{2}=(4,8)$

To find the radius, use the distance formula to find the distance from the center to one of the endpoints.

The distance formula is;

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$r=\sqrt{(4-4)^2+(10.5-8)^2}$

$r=\sqrt{0^2+(2.5)^2}$

$r=\sqrt{0^2+(2.5)^2}=2.5$

The equation of the circle in standard form is given by;

$(x-h)^2+(y-k)^2=r^2$

We substitute the center and the radius into the formula to get;

$(x-4)^2+(y-8)^2=2.5^2$

$(x-4)^2+(y-8)^2=6.25$

6 0

2 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago