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3 years ago

What is the value of x?

Mathematics

1 answer:

masya89 [10]3 years ago

6 0

Answer:

116 degrees

Explanation:

If we divide the triangle from the vertex of the triangle, the vertex angle of each triangle would lie on a straight line in the triangle. Angle on a straight line is 180 degrees.

If the vertex angle of one triangle is 64 degrees then the vertex angle x of the other triangle is 180 degrees - 64 degrees = 116 degrees.

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If the function is an exponential growth then the factor is

MariettaO [177]

I am confused by this question. But I will try to answer. Is it Exponential Decay?

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4 years ago

40% of oaty pop cereal boxes contain a prize hannah plans to keep buying cereal until she gets a prize what is the probability h

il63 [147K]

Answer:

I would say B.

Step-by-step explanation:

Since 40% of the cereal has a prize, if you use a cube from 1-6 and use 1-2 as the prizes (2 is 40% of 6), and roll it, it would correctly represent the possible solution.

3 0

3 years ago

Lim n-> infinity [1/3 + 1/3² + 1/3³ + . . . .+ 1/3ⁿ]

Verizon [17]

Answer:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]$

Let we first evaluate

$\rm :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} }$

Its a Geometric progression with

$\rm :\longmapsto\:a = \dfrac{1}{3}$

$\rm :\longmapsto\:r = \dfrac{1}{3}$

$\rm :\longmapsto\:n = n$

So, Sum of n terms of GP series is

$\rm :\longmapsto\:S_n = \dfrac{a(1 - {r}^{n} )}{1 - r}$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{1 - \dfrac{1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{3 - 1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{2}{3} } \bigg]$

$\bf\implies \:S_n = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

Hence, 

$\bf :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

Therefore, 

$\purple{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}\bigg[1 - 0 \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}$

Hence, 

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]} = \frac{1}{2}}}$

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<h3>Explore More</h3>

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}$

8 0

3 years ago

Nico, Aditi, Erick, Raj, and Sandra are solving the equation.

Alex777 [14]

Answer:

Nico, Raj and Sandra are correct

Step-by-step explanation:

we have

$\frac{2}{5}(5+x)=8$

Method 1

Solve for x

That means ----> isolate the variable x

we have

$\frac{2}{5}(5+x)=8$

Multiply by 5/2 both sides

$(\frac{5}{2})\frac{2}{5}(5+x)=(\frac{5}{2})8$

$5+x=20$

subtract 6 both sides

$5+x-5=20-5$

$x=15$

Method 2

Solve for x

That means ----> isolate the variable x

we have

$\frac{2}{5}(5+x)=8$

Multiply by 1/2 both sides

$(\frac{1}{2})\frac{2}{5}(5+x)=(\frac{1}{2})8$

$\frac{1}{5}(5+x)=4$

Multiply by 5 both sides

$(5)\frac{1}{5}(5+x)=(5)4$

$5+x=20$

subtract 6 both sides

$5+x-5=20-5$

$x=15$

Method 3

Solve for x

That means ----> isolate the variable x

we have

$\frac{2}{5}(5+x)=8$

Apply distributive property left side

$\frac{2}{5}(5)+\frac{2}{5}(x)=8$

$2+\frac{2}{5}(x)=8$

subtract 2 both sides

$2+\frac{2}{5}(x)-2=8-2$

$\frac{2}{5}(x)=6$

Multiply by 5/2 both sides

$(\frac{5}{2})\frac{2}{5}(x)=(\frac{5}{2})6$

$x=15$

therefore

Nico, Raj and Sandra are correct

6 0

4 years ago

Read 2 more answers

A rectangular vegetable patch has a perimeter of 28 meters and an area of 48 square

aliina [53]

Answer:

Length = 8 meters

Width = 6 meters

Step-by-step explanation:

Given the following information:

Perimeter of a rectangular vegetable patch: 2(L + W) = 28 meters

Area = L × W = 48 m²

We can solve for the dimensions by using both equations.

First, let's use the formula for the perimeter of the rectangular vegetable patch, and isolate one of the given variables:

2(L + W) = 28

Divide both sides by 2:

$\frac{2(L + W)}{2} = \frac{28}{2}$

L + W = 14

Subtract W from both sides to isolate L:

L + W - W = 14 - W

L = 14 - W

Next, we'll take the formula for the area, and substitute the value for the L from our previous step:

A = 48 = L × W

48 = W × (14 - W)

Distribute W into the parenthesis:

48 = 14W - W²

Add W² and subtract 14W to both sides:

W² - 14W + 48 = 14W - 14W - W² + W²

W² - 14W + 48 = 0 ⇒ This represents a quadratic equation in standard form. We can use the coefficient and constant values to solve for its roots.

a = 1, b = -14, and c = 48

Substitute these values into the quadratic equation:

$x = \frac{-b +/- \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{14 +/- \sqrt{(-14)^{2}-4(1)(48)}}{2(1)}$

$x = \frac{14 +/- \sqrt{196-192}}{2}$

$x = \frac{14 +/- \sqrt{4}}{2}$

$x = \frac{14 + 2}{2}$ , $x = \frac{14 - 2}{2}$

x = 8, x = 6

Now, we can substitute these values into the formulas for the perimeter and area to find the true dimensions of the rectangular vegetable patch.

Perimeter: 2(L + W) = 2(8 + 6) = 28 meters

Area = L × W = 8 × 6 = 48 m²

Therefore, the dimensions of the rectangular vegetable patch are:

Length = 8 meters

Width = 6 meters

Dimensions: 8 × 6 meters.

3 0

3 years ago