The matrix left. vphantom{ begin{array}{c} ! ! ! end{array}} right]left. vphantom{ begin{array}{c} ! ! ! end{array}} right[ boldsymbol{ vec{v}_2} = left. vphantom{ begin{array}{c} ! ! ! end{array}} right]and left. vphantom{ begin{array}{c} ! ! ! end{array}} right[ boldsymbol{ vec{v}_1} = lambda_2 = 3. find eigenvectors corresponding to these eigenvalues. lambda_1 = -9 and displaystyle left lbrack begin{array}{rr} -15 & -9 12 & 9 end{array} right rbrackhas eigenvalues

the circle one

it is true

step-by-step explanation:

all the combinations add up i used   calculator

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.

f(x)=4

step-by-step explanation: