How do i find lines AC, BD, and EC?

Answer: B: 4

Step-by-step explanation:

A number is divisible by 15 if the number ends on a 0 or a 5, and the sum of its digits is divisible by 3.

We want to find a 5 digit flippy number, this can be: modeled with:

N = ababa

Where a and b are single digit numbers.

If we impose that N must end with a zero, then we will have:

N  = 0b0b0 = b0b0

This is a 4-digit number, so we can discard all the options that end with a zero.

Then the only option that we have are the numbers like:

B = 5b5b5

This number will be only divisible by 15 if:

5 + b + 5 + b + 5 = K is divisible by 3.

Then let's try find the possible values of b.

K = 3*5 + 2*b

K has two terms, the left term is already divisible by 3, then K will be divisible by 3 only if the other term is also divisible by 3.

Then we want 2*b to be divisible by 3.

And 2 is a prime number, then b must be divisible by 3, and we know that b is a number between {0,1 , 2, 3, 4, 5, 6, 7, 8 ,9}

The options are:

2*0 = 0 is divisible by 3.

2*3  = 6 is divisible by 3.

2*6 = 12 is divisible by 3

2*9 = 12 is divisible by 3.

Then we have four values of b such that:

N = 5b5b5 is divisible by 15.

Then the correct option is: B: 4

Answer:

y=-2x+1

Step-by-step explanation:

y-y=m(x-x1)

y-(-3)=-2(x-2)

y+3=-2x+4

-3 -3

y=-2x+1

1. they both have 2 different congruent sides

2. angles ACD and BEC, since they are the same triangles, and are at congruent ends

3. since they are all the same triangle and has just been rotated some, you can figure out that it is the same triangle. once you figure out the angles for one of the triangles, you can figure outit for the other 2 since they are all congruent.

i hope this helped :)

Answer:

relax jamal dont pull up the 9

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

There is a common ratio r between consecutive terms in the sequence, that is

- 6 ÷ - 3 = - 12 ÷ - 6 = - 24 ÷ - 12 = 2

This indicates the sequence is geometric with n th term

[tex]a_{n}[/tex] = a₁ [tex]r^{n-1}[/tex]

where a₁ is the first term and r the common ratio

Here a₁ = - 3 and r = 2 , thus

a₁₁ = - 3 × [tex]2^{10}[/tex] = - 3 × 1024 = - 3072 → C

Answer:

y-5=1/3(x-3)

y=1/3x+4

no standard form

Step-by-step explanation:

Answer:

33b - 8

Step-by-step explanation:

You solve for the perimeter like you would for any other shape by adding all the sides

The perimeter is side A + side B + side C.

so it's 9b+8 + 12b-8 + 12b-8

Add all the like terms

9b + 12 b + 12b = 33b

8 - 8 - 8= -8

= 33b -8

Answer:

4x+13

Step-by-step explanation:

bc 12-8=4

6+7÷13

Responder:

225 kilometros

Explicación paso a paso:

Dado que :

Velocidad de funcionamiento = 30 km / h

Duración = 45 minutos

Número de días = 10 días

Distancia total recorrida por día:

Duración en horas:

45 minutos / 60 = 0,75 hora (s)

Así, la distancia recorrida por día:

Tiempo de velocidad

30 km / h * 0,75 h = 22,5 km

Por lo tanto, la distancia total recorrida durante 10 días:

Distancia recorrida diariamente * 10

22,5 kilometros * 10 = 225 kilometros

Answer:

The coordinates of B are: (a, b)=(-2, -6)

Step-by-step explanation:

Let (a, b) be the coordinates of the point 'B'.

As

The midpoint of AB is M(-4, 0)

The coordinates of A are (-6, 6)

The midpoint (x,y) of line joining points (x₁, x₂) and (y₁, y₂)

[tex]x=\frac{x_1+x_2}{2},\:y=\frac{y_1+y_2}{2}[/tex]

-4 = (-6+a)/2  and  0 = (6+b)/2

-8=-6+a          

-8+6=a                  

a=-2

0=(6+b)/2

0=6+b

b=-6

Therefore, the coordinates of B are: (a, b)=(-2, -6)

Answer:

BCE

Step-by-step explanation:

I hoped this helped and mark me as brainly pls

Answer:

a c b d

Step-by-step explanation:

Answer:

S = {ABC, ABD, ACD, BCD}

Step-by-step explanation:

We are told that Ariana, Boris, Cecile, and Diego are students in the service club.

Let's denote their names with letters;

Ariana = A

Boris = B

Cecile = C

Diego = D

We are told that Three of the four students will be chosen to attend a conference. This means that the sample space will be a set showing combination of any 3 out of the 4 students.

Thus, sample space is;

S = {ABC, ABD, ACD, BCD}

Answer:

A

Step-by-step explanation:

Answer:

Need more information, will get back to you.

Step-by-step explanation:

How many costumes does she want to make?

Answer:

hi! i'm so sorry about your headache, hope it gets better !

still need help?

Step-by-step explanation:

Answer: 100. Can I get a Brainlest.

100 is the answer

Multiply.....

God Bless

Answer:

i) [tex]P(X=12)=0.0355[/tex]

ii) [tex]P(X<10)=0.7553[/tex]

iii) [tex]P(X\geq 5)=0.9490[/tex]

Step-by-step explanation:

Let's start by defining the random variable ⇒

[tex]X:[/tex] '' Number of students that will get 7 marks out of 10 in first attempt of a certain Quiz while attempting Quiz on BlackBoard LMS ''

[tex]X[/tex] is a discrete random variable.

The probability of a randomly selected student getting 7 marks out of 10 is 0.4

(This is a data from the question).

Now, if we assume independence between the students while they are doing the Quiz and also we assume that this probability remains constant , we can modelate [tex]X[/tex] as a binomial random variable ⇒

[tex]X[/tex] ~ Bi (n,p)

Where ''n'' and ''p'' are the parameters of the variable.

''n'' is the number of students attempting the Quiz and ''p'' is the probability that a student will get 7 marks out of 10 which is 0.4 ⇒

[tex]X[/tex] ~ Bi (20, 0.4)  in the question.

The probability function for [tex]X[/tex] is

[tex]p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}[/tex]    (I)

Where [tex]\left(\begin{array}{c}n&x\end{array}\right)[/tex] is the combinatorial number define as

[tex]\left(\begin{array}{c}n&x\end{array}\right)=\frac{n!}{x!(n-x)!}[/tex]

Replacing the parameters in the equation (I) ⇒

[tex]P(X=x)=\left(\begin{array}{c}20&x\end{array}\right)(0.4)^{x}(0.6)^{20-x}[/tex]    (II)

For i) we need to find [tex]P(X=12)[/tex]

Then, we only need to replace by [tex]x=12[/tex] in equation (II) ⇒

[tex]P(X=12)=\left(\begin{array}{c}20&12\end{array}\right)(0.4)^{12}(0.6)^{8}=0.0355[/tex]

For ii) we need to calculate [tex]P(X<10)[/tex]

This probability is equal to ⇒ [tex]P(X<10)=P(X\leq 9)[/tex] and to calculate it we need to sum [tex]P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)[/tex]

[tex]+P(X=6)+P(X=7)+P(X=8)+P(X=9)[/tex]

We can do it summing each term or either using any program.

The result is [tex]P(X<10)=P(X\leq 9)=0.7553[/tex]

Finally for iii) we need to find  ⇒ [tex]P(X\geq 5)[/tex]

This probability is equal to ⇒ [tex]P(X\geq 5)=1-P(X\leq 4)[/tex] ⇒

[tex]1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)][/tex]

Again we can find each term by using the equation (II) or either using a program. The result is [tex]P(X\geq 5)=0.9490[/tex]

Answer:

u can use substitute or eliminate

Answer:

7

Step-by-step explanation:

65° + 15x + 10 = 180° (co-interior angles are equal)

15x + 75° = 180°

15x = 180° - 75°

15x = 105°

x = 105/15

x = 7

Answer:

9.28

Step-by-step explanation:

Answer:

a0.50 + b0.30 ≤ $3.00

Step-by-step explanation:

From the question, we have:

Gumballs cost per unit = $0.50

Jawbreakers cost = $0.30

Total budget of Bob = $3.00

Let a represents the number of Gumballs Bob can buy.

Let b represents the number of jawbreakers Bob can buy.

Therefore, inequality that represents the number of gumballs and jawbreakers Bob can buy can be given as follows:

a0.50 + b0.30 ≤ $3.00

The inequality implies that total amount Bob can spend on both gumballs and jawbreakers can be less or equal to $3.00 but cannot be greater than $3.00.

Answer:

HI FG

Step-by-step explanation:

Answer:

34

Step-by-step explanation:

8/2 = 4

6 times 5 = 30

30 + 4 = 34

Answer:

No, it doesn't.

Step-by-step explanation:

Whenever there's a "th" at the end of a mathematical word, it means decimal place. A tenth is the first number in a decimal. Ex. (Underlined is the tenth) 1.27. If you round them to the tenth, you would get 20 and 30. Therefore 20 does not round to 30 if rounding to the tenths.

Answer:

It would actually be 20.

Step-by-step explanation:

When rounding to the nearest tenth, you need to pay attention to the  second-to-last digit in the number. If the number is anything from 0 - 4 it rounds down. If the number is anything from 5 - 9, it rounds up.

Feel free to give brainliest.

Have an excellent day!

Answer:

The 3, and -4 are the places where you start, then just graph 1/2 and -3 as slope

Step-by-step explanation:

Step-by-step explanation:

you start at -4 and then you go down 3 and to the right 2

Answer:First suppose that the roots of the equation

x2−bx+c=0(1)

are real and positive. From the quadratic formula, we see that the roots of (1) are of the form

b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2.

For the root or roots to be real, we require that b2−4c≥0, that is, b2≥4c. In order for them to be positive, we require that

b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0.

This immediately tells us that b>0, but we can go further. We can rearrange this to get

b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,

which (assuming that b>0) is true if and only if

b2>b2−4c,

since both sides of the inequality are positive so we may square. But then

4c>0.

That is, if the roots are real and positive then b>0 and b2≥4c>0.

Now suppose that b>0 and b2≥4c>0.

Then the roots of (1) are real since b2−4c≥0, and b>0 guarantees that the root b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ is positive.

So it remains to show that b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0. We have that

4c>0,

so that

b2>b2−4c,

then square rooting shows that

b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,

so the roots of (1) are real and positive, as required.

Approach 2

the curve y equals x squared minus b x plus c showing two positive roots for y equals zero

This is intended to be a proof without words! We have from the diagram that:

If c, b and b2−4c are all positive, there are two real positive roots for x2−bx+c=0 (if b2=4c, we have two real positive equal roots).

If there are two real positive roots for x2−bx+c=0, then c and b are positive and b2−4c is non-negative.

(Why is the distance between the roots b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√?)

Approach 3

When solving problems about the roots of polynomials, it is often useful to find expressions those roots must satisfy and see if this tells us anything new. If α and β denote the roots of the equation, then

x2−bx+c=(x−α)(x−β)=x2−(α+β)x+αβ

and so α+β=b and αβ=c.

We also know that the roots of a quadratic equation are real if and only if the discriminant is non-negative, that is, if and only if b2−4c≥0.

Using these facts, if α and β are both real and positive, then b=α+β>0, c=αβ>0 and b2≥4c, as above.

Conversely, if b>0 and b2≥4c>0, then we know the discriminant is positive and hence both roots are real. We also have that

αβ>0(2)

and

α+β>0.(3)

As α and β are both real, by (2), we know that α and β are either both positive or both negative. However, if α and β were both negative, then (3) could not possibly hold. Hence α and β are both positive, as required.

We now sketch on a graph the region where b>0, c>0 and b2≥4c:

The curve b squared = 4 c as a quadratic with the c-axis vertical and the b-axis horizontal. The region below it is shaded.

The region of the b-c plane for which b>0, c>0 and b2≥4c

Sketch the region of the b-c plane in which the roots of the equation are real and less than 1 in magnitude.

We know that in order for the roots to be real we need b2≥4c as in the first part. We now need to find the region where

−1<b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.(4)

We have b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2≤b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2 so we only need to consider the values for which both

−1<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2andb+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.

Firstly, we will consider the values for which

b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2.

Rearranging gives

b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2−b.

So b<2, as the square root is non-negative, and we can square both sides to get

b2−4c<4−4b+b2,

which we may rearrange to find c>b−1.

We will now consider the values for which

−2<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√.

Similarly, we can rearrange to get

b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<b+2.

So b>−2, and we can, as before, square to get

b2−4c<b2+4b+4,

and hence c>−b−1.

To sketch the graph, we start by considering the boundary curves b2=4c, c=b−1 and c=−b−1, and the points at which they intersect. We can see that the two lines only intersect when b=0 and c=−1, and the lines intersect the curve when

b2=4b−4andb2=−4b−4

which rearrange to

(b−2)2=0and(b+2)2=0.

This tells us that each line intersects with the curve in only one place and so these lines must be tangent.

Is there a way we could have deduced this directly from (4)?

What do the lines being tangent signify in terms of our equation x2−bx+c=0?

Is this a representation of a well-known property of these equations?

Sketching the graph then yields the following picture:

The graph with the previous curve and the lines 4 c = 4 b minus 4 and 4 c = minus 4 b - 4. Each line touches the curve once and the two lines intersect at (0, minus 1). The region between the three lines/curves is shaded.

The shaded region is where b2≥4c, c>b−1, c>−b−1 and −2<b<2

We might notice that when we derived the inequalities c>b−1 and c>−b−1, a lot of the work we did was very similar.

We might also notice that the graph above is symmetric about the c axis. Why is this?

Does the graph give us any ideas about how we might deduce one inequality from the other?

Step-by-step explanation: