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2 years ago

What is (5,1) reflected on the y axis

Mathematics

2 answers:

9966 [12]2 years ago

7 0

Answer:

5 is how many units on the x axis (the horizontal one) and 1 is how many units on the y axis (the vertical one). So, the correct answer is 1.

asambeis [7]2 years ago

3 0

Answer:

(-5 , 1)

Step-by-step explanation:

If you are reflecting over the x-axis, you are changing the sign of the y.

If you are reflecting over the y-axis, you are changing the sign of the x.

In this case, you have the point (5 , 1). You are reflecting over the y-axis, which means that you are flipping the sign of the x value.

(5 , 1) reflected over the y-axis is (-5 , 1)

(-5 , 1)

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Hi what isthe answer

REY [17]

Answer:

2x^7

Step-by-step explanation:

2x^4 × x^3 =

= 2x^(4+3)

= 2x^7

8 0

2 years ago

a 9ft ladder is placed 4.5 feet from the base of a wall. How high up will the ladder reach? Round to the nearest hundredth.

spin [16.1K]

Answer:

$\huge\boxed{7.79 \ \text{feet}}$

Step-by-step explanation:

We know that we have a 9 foot long ladder resting upon a wall, with an unknown height, but the ladder is 4.5 feet away from the wall.

This can be represented as a triangle, where x is the unknown side.

| \

x | \ 9

| \

|_____\

4.5

We can use The Pythagorean Theorem to find the value of x. The theorem states that $a^2+b^2=c^2$ , where a and b are the lengths of the legs and c is the length of the hypotenuse.

However, we already know the hypotenuse and one leg. Therefore we can substitute inside the equation to find the missing value.

$a^2 + 4.5^2 = 9^2\\\\a^2 + 20.25 = 81\\\\a^2 = 60.75\\\\a \approx 7.79$

Hope this helped!

7 0

2 years ago

If g(x)=2x+1 and h(x)=x-2 then find (g-h)(x)

qwelly [4]

(g-h)(x) = 2x+1 -(<span>x-2)
</span>(g-h)(x) = 2x+1 - x + 2
(g-h)(x) = x + 3

4 0

2 years ago

Help I need to correct this

Levart [38]

I believe B,C, and D is correct...

please vote my answer brainliest. thanks!

4 0

3 years ago

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

$\displaystyle x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}n$

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

4 0

3 years ago