To address this issue, the procedure below should be utilized:
1- The Law of sines can be applied as follows:
The unknown angle can be determined by: 180°-65°-45°=70°
- The <span>distance Jake travels to the store is:
30/Sin(70°)=x/Sin(45°)
x=22.57
- The distance Maddie travels to the store is:
</span>30/Sin(70°)=y/Sin(65°)
<span> y=28.93
- The difference calculated is: 6.36
Thus, the result shows that the difference is 6.36 m.</span>
The answer
the full question is
If A(x1, y1), B(x2, y2), C(x3, y3), and D(x4, y4) create two line segments, and AB ⊥ CD, what condition must be satisfied to establish that AB ⊥ CD?
Let A(x1, y1) and B(x2, y2) represent the first line, while C(x3, y3) and D(x4, y4) represent the second line.
The slope for the first line is given by m = (y2 - y1) / (x2 - x1).
For the second line, the slope is m' = (y4 - y3) / (x4 - x3).
The necessary condition to demonstrate that AB ⊥ CD is
(y2 - y1) * (y4 - y3)
m × m' = --------- × ------------ = -1
(x2 - x1) (y4 - y3)
¡Hola! Bienvenido a!
Vamos a sumar cuántas canicas tenemos en total.
12+11+17+5=45
Queremos hallar la probabilidad de elegir una canica que no sea azul. Observemos cuántas canicas no son azules.
12+11+5=28
Tendremos esta probabilidad sobre 48.
28/48
Al simplificar, obtenemos 7/12 o alrededor de 58.33%.
¡Espero que esto ayude!
The diagrams for parts A and C are included here. For part B, we have circle O. We begin by drawing two radii OA and OC, connecting points A and C to create chord AC. The radius intersects chord AC at point B, bisecting AC into equal segments AB and BC. This gives us two triangles, ΔOBA and ΔOBC, where OA equals OC (since they're radii), OB equals OB (by the reflexive property), and AB is equal to BC (as stated in the question). By applying the SSS triangle congruence criterion, we conclude that ΔOBA is congruent to ΔOBC, allowing us to deduce that ∡OBA equals ∡OBC, both measuring 90°. Thus, OB is perpendicular to AC. Moving on to part D, we again work with circle O and draw the two radii OA and OC, joining points A and C to create chord AC. The radius intersects AC at point B, where AB is perpendicular to AC, meaning ∡B equals 90°. We then consider the right triangles ΔOBA and ΔOBC, and given OA equals OC (the radii), and OB equals OB (reflexive property), we conclude through the HL triangle congruence that ΔOBA is congruent to ΔOBC. Consequently, we find BA equal to BC, thus OB bisects AC.
Answer:
Step-by-step explanation:
Review the provided matrix
![A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%26-2%263%5C%5C2%2617%260%5C%5C3%2622%268%5Cend%7Barray%7D%5Cright%5D)
Let matrix B be defined as
![B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db_%7B11%7D%26b_%7B12%7D%26b_%7B13%7D%5C%5Cb_%7B21%7D%26b_%7B22%7D%26b_%7B23%7D%5C%5Cb_%7B31%7D%26b_%7B32%7D%26b_%7B33%7D%5Cend%7Barray%7D%5Cright%5D)
It is stated that
![\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9%26-2%263%5C%5C2%2617%260%5C%5C3%2622%268%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Db_%7B11%7D%26b_%7B12%7D%26b_%7B13%7D%5C%5Cb_%7B21%7D%26b_%7B22%7D%26b_%7B23%7D%5C%5Cb_%7B31%7D%26b_%7B32%7D%26b_%7B33%7D%5Cend%7Barray%7D%5Cright%5D)
By comparing the corresponding elements from both matrices, we derive
Consequently, the needed values are .
