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3 years ago

Need help quick please thanks

Mathematics

1 answer:

Sloan [31]3 years ago

8 0

Start by FOILing: First, Outer, Inner, Last.

3x•x= 3x^2; 3x•2= 6x; 2•x= 2x; 2•2= 4

Put it all together

3x^2 + 6x + 2x + 4

Combine like terms to get your answer

3x^2 + 8x + 4

You might be interested in

(-5,3);5×=4y+9 can you help me solve

kompoz [17]

Answer:

-25 ≠ 21

Step-by-step explanation:

it looks like the question is asking you to solve 5x=4y+9 when given the ordered pair (-5,3)

you would substitute -5 for x and 3 for y in the equation

5(-5) = 4(3) + 9

-25 = 12 + 9

-25 = 21

the question is asking you to see if the left side is equal to the right side of the equation

-25 is not equal to 21, therefore (-5,3) is not a solution to the equation

4 0

3 years ago

I need help with this geometry question

MArishka [77]

Answer:

Approximately 7.8 units

Step-by-step explanation:

To find the distance between any two points, we can use the distance formula:

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

From the graph, we can see that A is (0,0). Let's let this be x₁ and y₁.

B is (-5,6). Let's let this be x₂ and y₂. So, substitute:

$d=\sqrt{(-5-0)^2+(6-0)^2}$

Simplify:

$d=\sqrt{(-5)^2+(6)^2}$

Square:

$d=\sqrt{25+36}$

Add:

$d=\sqrt{61}$

Take the square root. Use a calculator:

$d\approx7.8$

So, the distance between them is about 7.8 units.

And we're done!

7 0

3 years ago

dem82 [27]

Answer:

D) gallery: g

balcony: 2g

main floor: 2g + 225

Step-by-step explanation:

3 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let <em>X</em> denote the event that the student has an “on” day, and let <em>Y</em> denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

2 years ago

Can someone please help me this is geometry I don’t understand I have like more than an hour trying to figure this but I cancele

Aleks04 [339]

Answer:

5, last option

Step-by-step explanation:

Sides to be calculated are equal as both are opposite same angles, are adjacent to common hypotenuse of right triangles

3x+1=2x+6
x=6-1
x=5

6 0

3 years ago