Ask question

Ask question

All categories

erastovalidia [21]

2 years ago

14

Help Me on this!!

%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20" id="TexFormula1" title=" \: \: \: \: \: \: \: \: \: \: \: \: " alt=" \: \: \: \: \: \: \: \: \: \: \: \: " align="absmiddle" class="latex-formula">

1 answer:

cricket20 [7]2 years ago

7 0

$\qquad\qquad\huge\underline{{\sf Answer}}$

Here's the solution :

9. Statement : $\tt{VB\:\: bisects\;\; \angle EVO}$

Reason : $\tt Given$

10. Statement : $\tt \angle3 \cong \angle 4$

Reason : $\tt Definition \;\; of \;\; angle \;\; bisector$

11. Statement $\tt{VB\:\: bisects\;\; \angle EBO}$

Reason : $\tt Given$

12. Statement : $\tt \angle1 \cong \angle 2$

Reason : $\tt Definition \;\; of \;\; angle \;\; bisector$

13. Statement : $\tt \overline{BV }\cong \overline{BV}$

Reason : $\tt Common\:\: side$

14. Statement : $\tt \triangle \:BEV \cong \triangle \:BOV$

Reason : $\tt ASA \:\: postulate$

15. Statement : $\tt \angle E \cong \angle O$

Reason : $\tt Corresponding\;\;parts\:\: of\:\: Congruent \;\;Triangles$

You might be interested in

the length of the sides of a square are initially 0 cm and increase at a constant rate of 11 cm per second. suppose the function

Likurg_2 [28]

The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about The function of the area of the square is A(t)=121 $t^{2}$

Given that The length of a square's sides begins at 0 cm and increases at a constant rate of 11 cm per second. Assume the function f determines the area of the square (in cm2) given several seconds, t since the square began growing and asked to find the function of the area

Lets assume the length of side of square is x

$\frac{dx}{dt} =$ 11 $\frac{cm}{sec}$

⇒x=11t

Area of square= $(length of side)^{2}$

Area of square= $(11t)^{2}$ {as the length of side is 11t}{varies by time}

Area of square=121 $t^{2}$

Therefore,The function of the area of the square is A(t)=121 $t^{2}$

Learn more about area here:

brainly.com/question/27683633

#SPJ4

3 0

1 year ago

I need help ASAP due at 11:59

algol [13]

Answer:

Step-by-step explanation:

8 0

3 years ago

HELP with these questions

zlopas [31]

<u>Step-by-step explanation:</u>

transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

0 = ln (x - 4)

$e^{0} = e^{ln (x - 4)}$

1 = x - 4

<u> +4 </u> <u> +4 </u>

5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

1 = ln (x - 4)

$e^{1} = e^{ln (x - 4)}$

e = x - 4

<u> +4 </u> <u> +4 </u>

e + 4 = x

6.72 = x

(6.72, 1)

Domain: x - 4 > 0

<u> +4 </u> <u>+4 </u>

x > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

= - 3⁰⁺⁵

= - 3⁵

= -243

Horizontal Asymptote: y = 0 <em>(explanation above)</em>

5 0

3 years ago

Answer quickly please <3

sertanlavr [38]

Answer:

x is greater than -5

Step-by-step explanation:

3 0

3 years ago

Simplify<br> 53<br> Enter your answer in the box

vesna_86 [32]

Answer:

we need more information

Step-by-step explanation:

7 0

2 years ago