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2 years ago

Help I need the awnser asap

Mathematics

1 answer:

Mrac [35]2 years ago

4 0

Answer:

Is it not just EF?

Step-by-step explanation:

I mean BC is the bottom of the triangle so the corresponding part is just the other triangles bottom which is EF if it was just angle B it'd be just angle E but it's not it's both B and C so it's E and F

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17) The measures of two sides of a triangle are (2x + 3y) and (5x – 2y). If the perimeter

elena-14-01-66 [18.8K]

Answer:

5x +4y

Step-by-step explanation:

The perimeter of a triangle is the sum of side lengths. That fact can be used to find z, the length of the third side.

(2x +3y) +(5x -2y) +z = 12x +5y . . . . sum of sides is perimeter

z = (12x +5y) -(7x +y) . . . . . . . . . . . . subtract (7x+y) from both sides

z = 5x +4y . . . . . . . . . collect terms; the length of the third side

The third side of the triangle is (5x +4y).

4 0

3 years ago

Find what b equals 36 - 2b = -5(7b + 6)

umka21 [38]

Answer:

b=−2

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

I need some help to find the equations for the questions. please help asap.

Dmitry [639]

5) Lets use x for the unknown middle side. Since all the sides have to equal to perimeter, we can set everything equal 43.

(x-3)+x+(2x-2)=43

That would be your equation.

6) Knowing what we know from the previous problem, we can set the equation to 27.

(x-3)+x+4(x-3)=27 (we can plug in x-3 to represent the first side)

7) Since all the sides have to equal to perimeter, we can set everything equal 25.

a=2b-2

c=b+3

(2b-2)+b+(b+3)=25

7 0

3 years ago

1. Solve the following simultaneous equations using the matrix method.

dmitriy555 [2]

(i) Use the formula for the determinant of a 2×2 matrix.

$\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad-bc$

$\implies \det(A) = \begin{vmatrix}4 & -3 \\ 2 & 5\end{vmatrix} = 4\times5 - (-3)\times2 = \boxed{26}$

(ii) The adjugate matrix is the transpose of the cofactor matrix of A. (These days, the "adjoint" of a matrix X is more commonly used to refer to the conjugate transpose of X, which is not the same.)

The cofactor of the (i, j)-th entry of A is the determinant of the matrix you get after deleting the i-th row and j-th column of A, multiplied by $(-1)^{i+j}$ . If C is the cofactor matrix of A, then

$C = \begin{pmatrix}5&-2\\3&4\end{pmatrix}$

Then the adjugate of A is the transpose of C,

$\mathrm{adj}(A) = C^\top = \boxed{\begin{pmatrix}5&3\\-2&4\end{pmatrix}}$

(iii) The inverse of A is equal to 1/det(A) times the adjugate:

$A^{-1} = \dfrac1{\det(A)} \mathrm{adj}(A) = \boxed{\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix}}$

(iv) The system of equations translates to the matrix equation

$A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\16\end{pmatrix}$

Multiplying both sides on the left by the inverse of A gives

$A^{-1}\left(A\begin{pmatrix}x\\y\end{pmatrix}\right)=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}$

$\left(A^{-1}A\right)\begin{pmatrix}x\\y\end{pmatrix}=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix} \begin{pmatrix}6\\16\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\boxed{\begin{pmatrix}3\\2\end{pmatrix}}$

4 0

2 years ago

The voltage in a circuit is the product of two factors, the resistance and the current. If the voltage is 6ir + 15i + 8r+20, fin

Alenkinab [10]

Answer:

resistance: (2r +5)
current: (3i +4)

Step-by-step explanation:

The factors of the given expression are ...

6ir +15i +8r +20 = (3i +4)(2r +5)

Which factor is current and which is resistance is not clear. Usually, resistance is referred to using the variable r, so we suppose the expressions are supposed to be ...

resistance: (2r +5)

current: (3i +4)

8 0

3 years ago