Given points B(−12,2), T(−8,10), A(−7,12), F(−6,10), and G(−2,2), which of the following proves that △BAG~△TAF?

Mathematics

1 answer:

Yanka [14]2 years ago

5 0

Answer:

By the Distance Formula, BA=5√5, TA=√5, A F=√5, and AG=5√5. Since, BA/TA = AG/A F = 5/1, and ∠A≅∠A by the Reflexive Property, △BAG∼△TAF by SAS ∼.

Hope this helps!

Step-by-step explanation:

You might be interested in

Write -8 in simplest radical form.

Nadya [2.5K]

2.8 or if rounded to the nearest tenth so 2.8√8

4 0

2 years ago

1. Simplify. 5–1(3–2)

konstantin123 [22]

Answer with explanation:

Using following law of indices

$1.x^a \times x^b=x^{a+b}\\\\2. (x^m)^n=x^{mn}\\\\ 3. x^{-n}=\frac{1}{x^n}\\\\4. \frac{x^m}{x^n}=x^{m-n}$

$1.5^{-1}\times 3^{-2}=\frac{1}{5}\times\frac{1}{3^2}\\\\=\frac{1}{5\times9}\\\\=\frac{1}{45}\\\\2. \frac{mn^{-4}}{p^0*q^{-2}}=\frac{mq^2}{n^4}\\\\ 3.0.0042=4.2\times 10^3\\\\4.6.12 \times 10^3=6.12 \times 1000=6120\\\\ 5.0.5 \times(8\times10^5})=\frac{1}{2}\times8 \times 10^5=4\times 10^5\\\\6. (9 \times 10^3)^2=9^2 \times (10^3)^2=81 \times 10^6\\\\7.( \frac{1}{2}a^{-4}\times b^2 )_{a=-2,b=4}{\text{or}}( \frac{1}{2}a^{4}\times b^{-2} )_{a=-2,b=4}=\frac{1}{2}\times (-2)^{-4}\times(4)^2{\text{or}}=\frac{1}{2}\times (-2)^{4}\times(4)^{-2}=\frac{16}{2\times16}=\frac{1}{2}$

$9. (4\times x\times y^2)^3\times (xy)^5=4^3\times x^3\times (y^2)^3\times x^5 \times y^5\\\\=64\times x^{3+5}\times y^{5+6}\\\\=64x^8y^{11}\\\\10.(2\times t^3)^3(0.4\times r)^2=2^3 \times (t^3)^3\times (0.4)^2 \times r^2=8\times t^9 \times 0.16 \times r^2=1.28r^2t^9$