-6t-7=17 what does t equal

Mathematics

1 answer:

Grace [21]3 years ago

7 0

Answer:

t=-4

Step-by-step explanation:

I know you don't care about any explanation, and just want the answer, but to get this, you can add 7 to both sides. Then you get -6t=24. Divide both sides by -6 and you get -4

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A triangular pyramid is formed from three right triangles as shown below.

Bess [88]

Answer:

FH = 108

Step-by-step explanation:

The given figure requires we use the Pythagorean theorem to write two relations involving right triangle side lengths. The Pythagorean theorem tells us the square of the hypotenuse is the sum of the squares of the other two sides.

<h3>Triangle EGH:</h3>

EG² = GH² +HE²

GH² = EG² -HE² = 53² -28² = 2025 . . . . . solve for GH², use given values

<h3>Triangle FGH:</h3>

FG² = GH² +FH²

FH² = FG² -GH² = 117² -2025 = 11664 . . . . solve for FH², use known values

FH = √11664 = 108 . . . . . take the square root

The length FH is 108.

6 0

3 years ago

Find the value of the determinant using the method of expansion by minors; expand on the third row

valkas [14]

For the matrix

$\begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}$

the determinant using the method of expansion by minors, expanding on the third row is:

$\det \begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}=a_{31}\det \begin{bmatrix}{a_{12}} & {a_{13}} & {} \\ {a_{22}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}-a_{32}\det \begin{bmatrix}{a_{11}} & {a_{13}} & {} \\ {a_{21}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}+a_{33}\det \begin{bmatrix}{a_{12}} & {a_{12}} & {} \\ {a_{22}} & {a_{22}} & {} \\ {} & {} & {}\end{bmatrix}$

Answer:

First, we compute the determinants of the minors:

$\begin{gathered} \det \begin{bmatrix}{0} & {4} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=0+4=4, \\ \det \begin{bmatrix}{1} & {4} & {} \\ {1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=3-4=-1, \\ \det \begin{bmatrix}{1} & {0} & {} \\ {1} & {-1} & {} \\ {} & {} & {}\end{bmatrix}=-1-0=-1. \end{gathered}$

Therefore: