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3 years ago

Help? Please? I do not get this.

Mathematics

1 answer:

olchik [2.2K]3 years ago

5 0

Length- 48:8 simplifies to 6.
Width- 30:5 simplifies to 6 also.
I believe your answer would be 1 because 6/6 is simplified to that.

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I want please help me as soon as possible I need help please help me

goldfiish [28.3K]

Answer:

none of the above

Step-by-step explanation:

8 0

4 years ago

35. Explain the difference between the reciprocal of x^2

Scorpion4ik [409]

Answer:

As you can see, the difference between the reciprocal of $x^2$ and the inverse of $x^2$ is that $\frac{1}{x^2} =x^{-2}$ and $\sqrt{x} =x^{\frac{1}{2}$ .

Step-by-step explanation:

First lets find both the reciprocal of $x^2$ and the inverse of

Recall that the reciprocal of a value is where you take a fraction and swap the places of the terms. In the case of $x^2$ , 1 is the denominator, so

$\frac{x^2}{1} =\frac{1}{x^2}$

To find the inverse of a function, you first need swap the locations of x and y in the equation

$y=x^2\\\\x=y^2$

Now, you need to solve for y

$y^2=x\\\\y=\sqrt{x}$

Now, lets rewrite each of these to better compare them

$\frac{1}{x^2} =x^{-2}\\\\\sqrt{x} =x^{\frac{1}{2}$

As you can see, the difference between the reciprocal of $x^2$ and the inverse of

7 0

3 years ago

36 2/5 x what = 109 1/5

tamaranim1 [39]

I am not sure what the answer is but covert them into a decimal and then divide it

6 0

3 years ago

Read 2 more answers

What is the answer I need to know

kolezko [41]

K = -18 to check, -4 + k/3 = -10 is the equation. -18/3 = -6, and -4 + -6 = -10. Mark as branliest if you can. :)

3 0

3 years ago

9.4 The heights of a random sample of 50 college stu- dents showed a mean of 174.5 centimeters and a stan- dard deviation of 6.9

gladu [14]

Answer: a) (176.76,172.24), b) 0.976.

Step-by-step explanation:

Since we have given that

Mean height = 174.5 cm

Standard deviation = 6.9 cm

n = 50

we need to find the 98% confidence interval.

So, z = 2.326

(a) Construct a 98% confidence interval for the mean height of all college students.

$x\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=(174.5\pm 2.326\times \dfrac{6.9}{\sqrt{50}})\\\\=(174.5+2.26,174.5-2.26)\\\\=(176.76,172.24)$

(b) What can we assert with 98% confidence about the possible size of our error if we estimate the mean height of all college students to be 174.5 centime- ters?

Error would be

$\dfrac{\sigma}{\sqrt{n}}\\\\=\dfrac{6.9}{\sqrt{50}}\\\\=0.976$

Hence, a) (176.76,172.24), b) 0.976.

8 0

3 years ago