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vredina [299]
3 years ago
14

What is the scale factor of figure A to figure B?

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0
B is 1/4 the size is of triangle A. The corresponding sides have a 4:1 ratio.
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Solve the following absolute value inequality. 3x - 727 x < 16 X > [?]​
RUDIKE [14]

Answer:

-2

Step-by-step explanation:

3|x-7|<= 27

then |x-7|<=27/3=9

hence, -9<=x-7<=9

therefore, -2<=x<=16

4 0
3 years ago
How to solve logarithms
statuscvo [17]
To solve<span> a logarithmic equation, rewrite the equation in exponential form and </span>solve<span>for the variable</span>
5 0
3 years ago
Using the problem below, solve and explain how you got the solution to the system of equation.
Paul [167]

Answer:

(5, -3)

y= -3

x= 5

Step-by-step explanation:

-x-2y=1

x+y=2

solve the equation

-x-2y=1

x=2-y

substitute the value of x into an equation

-(2-y)-2y=1

remove parenthesis

-2-y-2y=1

subtract y to 2y

-2-y=1

add 2 from both sides

-y=3

divide both sides by -y

y= -3

substitute the value of y into an equation

x=2-(-3)

remove parenthesis

x=2+3

add 2 to 3

x=5

----------

(5,-3)

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7 0
3 years ago
Course challenge
krek1111 [17]

Answer:

x=36

Divide the corresponding side (16) by 4/7

3 0
2 years ago
Read 2 more answers
Solve with cramer's rule x+2y+3z=11, 2x+y+2z=10, 3x+2y+z=9
nalin [4]

Answer:

x = 2 , y = 0 , z = 3

Step-by-step explanation:

Cramer's rule is a rule through which we can find the solution of linear equation.

we have the three linear equations as

x+2y+3z=11

2x+y+2z=10

3x+2y+z=9

AX=B  

A: coefficient matrix

X= unknown vectors(x,y,z)

D = values of the linear equation (11 , 10 , 9)

now we find the determinant of the given linear equation

determinant of the matrix will be

A = \left[\begin{array}{ccc}1&2&3\\2&1&2\\3&2&1\end{array}\right]  = 1(1-4) - 2(2-6) + 3(4 - 3)

                    = 1(-3) - 2(-4) + 3(1)

                    = -3+8+3 = 8

also D\neq 0

so the determinant is Non zero we can apply Cramer's rule

we will be replacing the first column of the coefficient matrix A with the values of D

by replacing the first column we will get the value of the variable 'x'

Dx =  \left[\begin{array}{ccc}11&2&3\\10&1&2\\9&2&1\end{array}\right]   = 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16

x = \frac{Dx}{D}  = \frac{16}{8} = 2

similarly

Dy = \left[\begin{array}{ccc}1&11&3\\2&10&2\\3&9&1\end{array}\right] = 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0

y = \frac{Dy}{D} = 0

Dz= \left[\begin{array}{ccc}1&2&11\\2&1&10\\3&2&9\end{array}\right] = 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24

z = \frac{Dz}{D} = \frac{24}{8} = 3

so we have the solution as

x = 2 , y = 0 , z = 3

Therefore the solution for the given linear equations is (2,0,3).

3 0
3 years ago
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