Helllpppp. don’t answer if you don’t know

Help Please

yaroslaw [1]

Answer:

Its (10,∞)

Step-by-step explanation:

Trust me I just took the exam and got it correct

5 0

3 years ago

(5x – 3y) (3x + 4y)

MariettaO [177]

Answer:

$\boxed{\red{15 {x}^{2} + 11xy - 12 {y}^{2}}}$

Step-by-step explanation:

$\blue{(5x - 3y)(3x + 4y)} \\ \blue{5x(3x + 4y) - 3y(3x + 4y)} \\ \blue{15 {x}^{2} + 20xy - 9xy - 12 {y}^{2} } \\ \pink{= 15 {x}^{2} + 11xy - 12 {y}^{2} }$

6 0

3 years ago

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Choose the domain for which each function is defined

klemol [59]

Answer:

Part 1) $f(x)=\frac{x+4}{x}$ -----> $x\neq 0$

Part 2) $f(x)=\frac{x}{x+4}$ ----> $x\neq -4$

Part 3) $f(x)=x(x+4)$ ----> All real numbers

Part 4) $f(x)=\frac{4}{x^2+8x+16}$ ----> $x\neq -4$

Step-by-step explanation:

we know that

The domain of a function is the set of all possible values of x

Part 1) we have

$f(x)=\frac{x+4}{x}$

we know that

In a quotient the denominator cannot be equal to zero

so

For the value of x=0 the function is not defined

therefore

The domain is

$x\neq 0$

Part 2) we have

$f(x)=\frac{x}{x+4}$

we know that

In a quotient the denominator cannot be equal to zero

so

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

Part 3) we have

$f(x)=x(x+4)$

Applying the distributive property

$f*(x)=x^2+4x$

This is a vertical parabola open upward

The function is defined by all the values of x

therefore

The domain is all real numbers

Part 4) we have

$f(x)=\frac{4}{x^2+8x+16}$

we know that

In a quotient the denominator cannot be equal to zero

so

Equate the denominator to zero

$x^2+8x+16=0$

Remember that

$x^2+8x+16=(x+4)^2$

( $x+4)^2=0$

The solution is x=-4

so

For the value of x=-4 the function is not defined

therefore

The domain is

$x\neq -4$

6 0

3 years ago

A special deck of cards has ten cards. Four are green (G), four are blue (B), and two are red (R). When a card is picked, the co

a_sh-v [17]

Answer:

{GT, GH, BT, BH, RT, RH};

1 /5;

Mutually exclusive ;

Not mutually exclusive.

Step-by-step explanation:

Given :

Green cards, G = 4

Blue cards, B = 4

Red cards, R = 2

For a coin toss :

{H, T}

A card is picked, then a coin is tossed :

Sample space :

{GT, GH, BT, BH, RT, RH}

2.) probability of picking a green card, then probability of landing a head on a coin toss

P(A) = number of required outcome / Total possible outcomes

P(A) = P(green card) * P(Head)

P(A) = (4 / 10) * (1/2)

P(A) = 4 /20

P(A) = 1/5

C.)

Both A and B are mutually exclusive, since both event A and event B cannot occur together, since both red and green cannot be picked during a single pick, this either a red is picked or green is picked, then they are A and B are mutually exclusive.

D.) Event A and C are not mutually exclusive, picking a green card, event A and picking a red or blue card, event B. Both event can happen simultaneously, hence, event A and B are not mutually exclusive.

6 0

3 years ago

Over 6.8 months, a tree grew 7.48 inches.

evablogger [386]

The tree's grow rate is 1.1 per month
7.48/6.8
=1.1

4 0

3 years ago

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