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3 years ago

Solve this equation.

Mathematics

2 answers:

cestrela7 [59]3 years ago

8 0

Answer:

OB=4

Step-by-step explanation:

Step 1: Subtract 5 from both sides.

4x+5−5=21−5

4x=16

Step 2: Divide both sides by 4.

4x/4=16/4

x=4

Answer:

x=4

Lena [83]3 years ago

7 0

Answer:

OB: 4

Step-by-step explanation:

4x+5=21

Next step is to subtract 5 on both sides to keep it an equation.

4x=21-5

4x=16

Finally, divide by 4 on both sides for x to be alone and for it to stay and equation.

x=4!

There you go! Hope this helped, this is my first time answering a question! :p

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julia-pushkina [17]

Answer:

702.1

Step-by-step explanation:

Use the formula for the diagonal of a cuboid.

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3 years ago

Consider the graph of the function f(x) = 25

trasher [3.6K]

Considering it's horizontal asymptote, the statement describes a key feature of function g(x) = 2f(x) is given by:

Horizontal asymptote at y = 0.

<h3>What are the horizontal asymptotes of a function?</h3>

They are the limits of the function as x goes to negative and positive infinity, as long as these values are not infinity.

Researching this problem on the internet, the functions are given as follows:

$f(x) = 2^x$ .

$g(x) = 2f(x) = 2(2)^x$

The limits are given as follows:

$\lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} 2(2)^x = \frac{2}{2^{\infty}} = 0$

$\lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} 2(2)^x = 2(2)^{\infty} = \infty$

Hence, the correct statement is:

Horizontal asymptote at y = 0.

More can be learned about horizontal asymptotes at brainly.com/question/16948935

#SPJ1

3 0

1 year ago

The equation of a parabola is y=x^2-10x+27. Write the equation in vertex form

azamat

Answer:

$\large \boxed{y = (x - 5)^{2} + 2 }$

Step-by-step explanation:

y = x² - 10x + 27

y = ax² + bx + c

This is the general form of the equation for a parabola.

We must convert it to the vertex form

y = (x - h)² + k, where (h,k) are the coordinates of the vertex.

We can do this by completing the square.

$\begin{array}{rcll}y & = & x^{2} - 10x + 27 & \\y - 27 & = & x^{2} - 10x & \text{Subtracted 27 from each side}\\y - 27&= & x^{2} - 10x + 25 - 25 & \text{Added and subtracted (b/2)}^{2}\\y - 27&= & (x - 5)^{2} - 25 & \text{Wrote the first three terms as the square of a binomial}\\y& = & \mathbf{(x - 5)^{2} + 2} & \text{Added 27 to each side}\\\end{array}\\\text{The vertex form of the parabola is $\large \boxed{\mathbf{y = (x - 5)^{2} + 2 }}$}$ The figure below shows that your parabola has its vertex at (5,2).