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Llana [10]
3 years ago
5

Is it A B C D Will be marked as crown

Mathematics
2 answers:
Solnce55 [7]3 years ago
3 0
The answer is B just took the test
galben [10]3 years ago
3 0
Actually I think that it is....
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The slope of a line is 1, and the y-intercept is -1. what is the equation of the line written in slope-intercept form?
elena-s [515]
The m in the equation is the slope and the y-intercept is the b.
y=1x-1
y=mx+b
4 0
3 years ago
8) Find the endpoint Cif M is the midpoint of segment CD and M (2, 4) and D (5,7)
Elenna [48]

Answer:

8. c. (-1, -1)

9. a. (-6, -1)

b. True

Step-by-step Explanation:

8. Given the midpoint M(2, 4), and one endpoint D(5, 7) of segment CD, the coordinate pair of the other endpoint C, can be calculated as follows:

let D(5, 7) = (x_2, y_2)

C(?, ?) = (x_1, y_1)

M(2, 4) = (\frac{x_1 + 5}{2}, \frac{y_1 + 7}{2})

Rewrite the equation to find the coordinates of C

2 = \frac{x_1 + 5}{2} and 4 = \frac{y_1 + 7}{2}

Solve for each:

2 = \frac{x_1 + 5}{2}

2*2 = \frac{x_1 + 5}{2}*2

4 = x_1 + 5

4 - 5 = x_1 + 5 - 5

-1 = x_1

x_1 = -1

4 = \frac{y_1 + 7}{2}

4*2 = \frac{y_1 + 7}{2}*2

8 = y_1 + 7

8 - 7 = y_1 + 7 - 7

1 = y_1

y_1 = 1

Coordinates of endpoint C is (-1, 1)

9. a.Given segment AB, with midpoint M(-4, -5), and endpoint A(-2, -9), find endpoint B as follows:

let A(-2, -9) = (x_2, y_2)

B(?, ?) = (x_1, y_1)

M(-4, -5) = (\frac{x_1 + (-2)}{2}, \frac{y_1 + (-9)}{2})

-4 = \frac{x_1 - 2}{2} and -5 = \frac{y_1 - 9}{2}

Solve for each:

-4 = \frac{x_1 - 2}{2}

-4*2 = \frac{x_1 - 2}{2}*2

-8 = x_1 - 2

-8 + 2 = x_1 - 2 + 2

-6 = x_1

x_1 = -6

-5 = \frac{y_1 - 9}{2}

-5*2 = \frac{y_1 - 9}{2}*2

-10 = y_1 - 9

-10 + 9 = y_1 - 9 + 9

-1 = y_1

y_1 = -1

Coordinates of endpoint B is (-6, -1)

b. The midpoint of a segment, is the middle of the segment. It divides the segment into two equal parts. The answer is TRUE.

4 0
4 years ago
9:15 in ratio simplest form
olchik [2.2K]

0.6:1

Hope it helps yah (◕ᴗ◕✿)

5 0
3 years ago
Read 2 more answers
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Fill in the table.<br>Input Values = 0, 2,4<br>Rule: subtract 3<br>Input<br>Output​
jarptica [38.1K]

Answer:

Input: 0, 2, 4

Output: 3, -1, 1

Step-by-step explanation:

0-3= -3

2-3= -1

4-3= 1

3 0
3 years ago
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