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3 years ago

Solve the equation. Then check your solution.

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

4 0

8x+5= 69

Move +5 to the other. Sign changes from +5 to -5.

8x+5-5= 69-5

8x= 64

Divide by 8

8x/8= 64/8

x= 8

Answer : x= 8

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For what value of x is the value of f(x)=4x-2 equal to 18

mina [271]

F(x) = 4(5) - 2 → f(x) = 20 - 2 → f(x) = 18

5 0

3 years ago

Read 2 more answers

Quadrilateral ABCD is inscribed in this circle.

Sunny_sXe [5.5K]

Answer: $m\angle A=116\°$

Step-by-step explanation:

The missing figure is attached.

For this exercise it is important to remember that, by definition, the opposite interior angles of an inscribed quadrilateral are supplementary, which means that their sum is 180 degrees.

Based on this, you can identify that the angle D and the angle B are opposite and, therefore, supplementary.

Knowing that, you can write the following equation:

$x+28\°=180\°$

Now you must solve for "x" in order to find its value. This is:

$x=180\°-28\°\\\\x=152\°$

Then:

$m\angle D=152\°$

You know that:

$m\angle A=(x-36)\°$

Therefore, since you know the value of "x", you can substitute it into $m\angle A=(x-36)\°$ and then you must evaluate, in order to find the measure of the angle A. This is:

$m\angle A=152\°-36\°\\\\m\angle A=116\°$

4 0

3 years ago

Describe the position of the function below relative to the graph of the indicated basic function.

matrenka [14]

From the given the option moved left 1 unit; reflected across the x axis is the correct.

what is function?

A function from a set X to a set Y in mathematics assigns each element of X exactly one element of Y. The set X is known as the function's domain, and the set Y is known as the function's codomain.

We have to describe the position of the function $f(x) = -2^x^+^1$ relative to the basic function $2^x$ .

Let the parent function is $2^x$ .

After shifting one unit right it becomes, $2^x^+^1$ .

After reflection in x axis it becomes, $-2^x^+^1$ .

Hence, from the given the option moved left 1 unit; reflected across the x axis is the correct.

To know more about function, click on the link

brainly.com/question/25638609

#SPJ1

7 0

1 year ago

At Grocery Mart, tomatoes are $2.75 per pound, or you can buy a 5

larisa [96]

Answer:

a) $2.75

b) $2.2

c) Second tomato option is better value option.

Step-by-step explanation:

First option to buy tomatoes = $2.75 per pound

Second option to buy tomatoes:

To buy a box which is a 5 pound box and the price is $11.

To find:

a) Unit price for the first tomato option.

b) Unit price for the second tomato option.

c) Better value option on the basis of price.

Solution:

Let the consider the unit here to be a pound.

So, we will find the price of tomato per pound to find the unit price of tomato.

a) Unit price of the first tomato option is : $2.75 per pound

b) Unit price of the second tomato option:

Price for 5 pounds = $11

Dividing both sides with 5 to find the price of 1 pound or unit price.

Price for $\frac{5}{5}$ = 1 pound = $\frac{\$11}5 = \$2.2$

c) To find the better value option, we can compare the unit prices of the two tomato options.

Here, we have the two tomato options with unit prices as:

$2.75 and $2.2

Second tomato option has a lower unit price, therefore, <em>second option</em> is better value option on the basis of price.

5 0

2 years ago

Find the taylor series centered at the given value of a and find the associated radius of convergence. (1) f(x) = 1 x , a = 1 (2

Kitty [74]

a) The radius of convergence is calculated as

R=1.

b) Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.

<h3>What is the associated radius of convergence.?</h3>

(a)

Take into consideration the function f with respect to the number a,

$f(x)=\frac{1}{x}, \quad a=1$

In case you forgot, the Taylor series for the function $f$ at the number a looks like this:

$\begin{aligned}f(x) &=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^{n} \\&=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{*}(a)}{2 !}(x-a)^{2}+\ldots\end{aligned}$

Determine the function f as well as any derivatives of the function $f by setting a=1 and working backward from there.

$\begin{aligned}f(x) &=\frac{1}{x} & f(1)=\frac{1}{1}=1 \\\\f^{\prime}(x) &=-\frac{1}{x^{2}} & f^{\prime}(1)=-\frac{1}{(1)^{2}}=-1 \\\\f^{\prime \prime}(x) &=\frac{2}{x^{3}} & f^{\prime \prime}(1)=\frac{2}{(1)^{3}}=2 \\\\f^{\prime \prime}(x) &=-\frac{2 \cdot 3}{x^{4}} & f^{\prime \prime}(1)=-\frac{2 \cdot 3}{(1)^{4}}=-2 \cdot 3 \\\\f^{(*)}(x) &=\frac{2 \cdot 3 \cdot 4}{x^{5}} & f^{(n)}(1)=\frac{2 \cdot 3 \cdot 4}{(1)^{5}}=2 \cdot 3 \cdot 4\end{aligned}$

At the point when a = 1, the Taylor series for the function f looks like this:

$f(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}+\cdots \\\\&=f(1)+\frac{f^{\prime}(1)}{1 !}(x-1)+\frac{f^{\prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime}(1)}{3 !}(x-1)^{3}+\cdots \\$

$&=1+\frac{-1}{1 !}(x-1)+\frac{2}{2 !}(x-1)^{2}+\frac{-2 \cdot 3}{3 !}(x-1)^{3}+\frac{2 \cdot 3 \cdot 4}{4 !}(x-1)^{4}+\cdots \\\\&=1-(x-1)+(x-1)^{2}-(x-1)^{3}+(x-1)^{4}+\cdots \\\\&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

In conclusion,

$&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

Find the radius of convergence by using the Ratio Test in the following manner:

$\begin{aligned}L &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \\&=\lim _{n \rightarrow \infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{(-1)^{n}(x-1)^{n}} \mid \\&=\lim _{n \rightarrow \infty}|x-1| \\&=|x-1|\end{aligned}$

The convergence of the series when L<1, that is, |x-1|<1.

The radius of convergence is calculated as

R=1.

For B

Take into consideration the function f with respect to the number a,

$a_{n}=(-1)^{n}(x-1)^{n}$

$f(x)=\left(x^{2}+2 x\right) e^{x}, a=0$ The Taylor series for f(x)=e^{x} at a=0 is,

$e^{2}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots$

$f(x) &=\left(x^{2}+2 x\right) e^{x} \\&=\left(x^{2}+2 x\right)\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+2 x\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right) \\&=x^{2}\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+\left(\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\frac{x^{6}}{4 !}+\ldots\right)+\left(2 x+2 x^{2}+\frac{2 x^{3}}{2 !}+\frac{2 x^{4}}{3 !}+\frac{2 x^{5}}{4 !}+\ldots\right) \\$

$&=\left(x^{2}+x^{3}+\frac{x^{4}}{2 !}\right) \\&=2 x+3 x^{2}+\left(1+\frac{2}{2 !}\right) x^{3}+\left(\frac{1}{2 !}+\frac{2}{3 !}\right) x^{4}+\left(\frac{1}{4 !}\right) x^{5}+\ldots \\&=2 x+3 x^{2}+2 x^{3}+\frac{5}{6} x^{4}+\frac{1}{4} x^{5}+\ldots$

Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.