All categories

3 years ago

Please answer correctly !!!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

aalyn [17]3 years ago

8 0

<h3>Answer: VT = 8</h3>

==========================================

Work Shown:

V is the midpoint of RT, so RV = VT

RV = 2x+4

VT = RV = 2x+4

RV+VT = RT ... by the segment addition postulate

RV+RV = RT ... replace VT with RV

2*(RV) = RT

2*(2x+4) = 8x .... plug in the given expressions

4x+8 = 8x

8 = 8x-4x

8 = 4x

4x = 8

x = 8/4

x = 2

Which means,

RV = 2x+4 = 2*2+4 = 8

VT = RV = 8

RT = 8x = 8*2 = 16

Note how

RV+VT = 8+8 = 16

which matches with the length of RT to help confirm our answer.

Ira Lisetskai [31]3 years ago

5 0

Answer:

$\boxed{line \: vt \: = 8.}$

Step-by-step explanation:

$if \: \boxed{ line\: rt} = \boxed{ line\: rv} + \boxed{ line\: vt} = 8x \\ then \: \boxed{ line\: rv} = \boxed{ line\: rt} = 2x + 4 \\ hence \to \\ 2(2x + 4) = 8x \\ 4x + 8 = 8x \\ 4x = 8 \\ \boxed{x = 2} \\ \\ if \: x = 2 :then \: \\ 2x + 4 = 2(2) + 4 = 8. \\ therefore \: \boxed{ line\: vt = 8.}$

♨Rage♨

You might be interested in

Rick has 7 cats. Carrie has 14 cats. How many fewer cats does Rick have

podryga [215]

Answer:

14-7 and 7

Step-by-step explanation:

8 0

3 years ago

Find the perimeter of Triangle ABC with the verticies A(-5,5) B(3,-5) and C(-5,1

finlep [7]

Hello!

Let's calculate a distance between two points using the Pythagorean theorem:

 $d ^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2$

Data:

$x_B = 3 

x_A = -5 

y_B = -5 

y_A = 5 

d_ {AB} =?$

Solving: distance from A to B

$d^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2 

d^ 2_ {AB} = (3 - (-5)) ^ 2 + (-5 - 5) ^ 2 

d^ 2_ {AB} = (8) ^ 2 + (-10) ^ 2 

d^ 2_ {AB} = 64 + 100 

d^ 2_ {AB} = 164 

d_{AB} = \sqrt {164} 

d_{AB} = \sqrt {2 ^ 2 * 41} 

\boxed {d_ {AB} = 2 \sqrt {41}}$

Solving:

distance from A to C

$d ^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2$

Data:

$x_C = -5
 
x_A = -5
 
y_C = 1 

y_A = 5 

d_ {AC} =?$

$d^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2 

d^ 2_ {AC} = (-5 - (-5)) ^ 2 + (1 - 5) ^ 2 

d^ 2_ {AC} = (0) ^ 2 + (-4) ^ 2 

d^ 2_ {AC} = 0 + 16 

d^ 2_ {AC} = 16 

d_ {AC} = \sqrt {16} 

\boxed {d_ {AC} = 4}
$

Solving:

distance from B to C

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2$

Data:

$x_C = -5 

x_B = 3 

y_C = 1 

y_B = -5 

d_ {BC} =?$

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2
 
d^ 2_ {BC} = (-5 - 3) ^ 2 + (1 - (-5)) ^ 2

d^ 2_ {BC} = (-8) ^ 2 + (6) ^ 2 

d^ 2_ {BC} = 64 + 36

d^ 2_ {BC} = 100 

d_ {BC} = \sqrt {100}

\boxed {d_ {BC} = 10}$

Now let's calculate the perimeter of the triangle ABC, knowing that the perimeter is the sum of the sides, then:

$p = d_ {AB} + d_ {AC} + d_ {BC} 

$

$p = 2\sqrt {41} + 4 + 10$

$
\boxed {\boxed {p = 14 + 2 \sqrt {41}}} \end {array}} \qquad \quad \checkmark$

8 0

3 years ago

Calculate the limit values:

Nataliya [291]

A) This particular limit is of the indeterminate form,
$\frac{ \infty }{ \infty }$
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

's
$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }$
So we take the derivatives and obtain,

$Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}$

Still it is of the same indeterminate form, so we apply the rule again,

$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}$

This simplifies to,

$Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0$

b) This limit is also of the indeterminate form,

$\frac{0}{0}$
we still apply the L'Hopital's Rule,

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }$

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }$

When we plug in zero now we obtain,

$Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1$
c) This also in the same indeterminate form

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }$

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }$

It is still of that indeterminate form so we apply the rule again, to obtain;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }$

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }$

This gives us;

$Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2$

d) $Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x$

For this kind of question we need to rationalize the radical function, to obtain;

$Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}$

We now divide both the numerator and denominator by x, to obtain,

$Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}$

This simplifies to,

$=\frac{2}{\sqrt{1+0}+1}=1$