All categories

2 years ago

In the figure, ∆BAT ≅ ∆CAT. Which statement is true by CPCTC?

Mathematics

1 answer:

agasfer [191]2 years ago

8 0

Answer:

BT=CT the first one because BAT=CAT and bc cut the similar vector between these two triangle

You might be interested in

Your boss sends you a text of a painting of the Yankee stadium.

kaheart [24]

Answer:

170.6 inches

The scale factor is an enlargement

Step-by-step explanation:

16 by 24 is the Painting by a scale factor of 2.25

A scale factor is an enlargment or reduction in a shape or quantity.

(16*24) is the size of the painting after Scale factor

16*24=384

Know get 384/2.25

170.666666667

170.6 inches

4 0

2 years ago

Find the common ratio, the 8th term, the explicit formula, the recursive formula, and the three terms in the sequence after the

skad [1K]

Answer:

I am also truly sorry

Step-by-step explanation:

6 0

3 years ago

(x-20) (2x+10) (x+40) find the value of x

katrin [286]

Answer:

2 • (x - 20) • (x + 5) • (x + 40)

Step-by-step explanation:

7 0

2 years ago

Find the distance between the point (-2, -3) and the line with the equation y=4-2/3x

otez555 [7]

Answer:

$D = \frac{25\sqrt{13}}{13}$

Step-by-step explanation:

Given

$y = 4 - \frac{2}{3}x$

$(x_1,y_1) = (-2,-3)$

Required

Determine the distance

$y = 4 - \frac{2}{3}x$

Write the above equation in standard form:

$Ax + By + C = 0$

So, we have:

$\frac{2}{3}x+y - 4 = 0$

By comparison:

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

The distance is calculated using:

$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Where:

$(x_1,y_1) = (-2,-3)$

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

This gives:

$D = \frac{|\frac{2}{3} * (-2) + 1 * (-3) - 4|}{\sqrt{(\frac{2}{3})^2 + 1^2}}$

$D = \frac{|-\frac{4}{3} -3 - 4|}{\sqrt{\frac{4}{9} + 1}}$

Take LCM

$D = \frac{|\frac{-4-9-12}{3}|}{\sqrt{\frac{4+9}{9}}}$

$D = \frac{|\frac{-25}{3}|}{\sqrt{\frac{13}{9}}}$

$D = |\frac{-25}{3}|/\sqrt{\frac{13}{9}}$

$D = \frac{25}{3}/\sqrt{\frac{13}{9}}$

Split the square root

$D = \frac{25}{3}/\frac{\sqrt{13}}{\sqrt{9}}$

Change / to *

$D = \frac{25}{3}*\frac{\sqrt{9}}{\sqrt{13}}$

$D = \frac{25}{3}*\frac{3}{\sqrt{13}}$

$D = \frac{25}{\sqrt{13}}$

Rationalize

$D = \frac{25}{\sqrt{13}} * \frac{\sqrt{13}}{\sqrt{13}}$

$D = \frac{25\sqrt{13}}{13}$

Hence, the distance is:

$D = \frac{25\sqrt{13}}{13}$

4 0

2 years ago

Solve for b. <br><br> 3a + 2b = c

Licemer1 [7]

B=(c-3a)/2
Should be the equation you get

5 0

3 years ago

Read 2 more answers