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3 years ago

What's the nth term of the following question:

Mathematics

1 answer:

Leona [35]3 years ago

5 0

Answer:

-21

Step-by-step explanation:

-13 +(-2) = -15

-15+(-2) = -17

-17+(-2 )=- 19

-19+ (-2)= -21

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What is the radical form of the expression 4 3/2?

mart [117]

Answer: D

Step by Step explanation:

6 0

3 years ago

Read 2 more answers

pretty sad and pathetic when half the answers on here are wrong and marked verified or brainliest. even more sad when 18/20 ques

tekilochka [14]

Answer:

Well im sure that it wasn't a persons intention it must be a mistake i have found this very useful and i have the basic

Step-by-step explanation:

5 0

3 years ago

Consider f (x) = StartRoot x squared minus 1 EndRoot and g (x) = StartRoot x squared + 1 EndRoot. What value(s) of x would make

dalvyx [7]

Answer:

Any value of x

Step-by-step explanation:

Given

$f(x) = \sqrt{x^2 - 1}$

$g(x) = \sqrt{x^2 + 1}$

Required

What value of x is $f(g(x)) = g(f(x))$

Solving for f(g(x))

$f(x) = \sqrt{x^2 - 1}$

$f(g(x)) = \sqrt{(\sqrt{x^2 + 1})^2 - 1}$

Solve the inner square

$f(g(x)) = \sqrt{(x^2 + 1 - 1}$

$f(g(x)) = \sqrt{x^2 } }$

$f(g(x)) = x$

Solving g(f(x))

$g(x) = \sqrt{x^2 + 1}$

$g(f(x)) = \sqrt{(\sqrt{x^2 - 1})^2 + 1}$

$g(f(x)) = \sqrt{x^2 - 1 + 1}$

$g(f(x)) = \sqrt{x^2 }$

$g(f(x)) = x$

Equate f(g(x)) and g(f(x))

$f(g(x)) = g(f(x))$

$x = x$

This implies that $f(g(x)) = g(f(x))$ at any value of x

8 0

4 years ago

The length of a rectangle is 5 inches more than the width. The area is 33 square inches. Find the length and width

Katen [24]

The length is 8.8 and the width is 3.8
Hope this helps!

5 0

3 years ago

(1 point) Let p be the joint density function such that p(x,y)=116xy in R, the rectangle 0≤x≤4,0≤y≤2, and p(x,y)=0 outside R. Fi

baherus [9]

Answer:

The answer is $\frac{7}{8}$ of the population.

Step-by-step explanation:

The question is wrong. The joint density function is $p(x,y)=(\frac{1}{16})xy$ in R

and $p(x,y)=0$ outside R.

R is defined as the rectangle $0\leq x\leq 4$ , $0\leq y\leq 2$

In order to find the fraction of the population satisfying the constraint $x\geq y$ , we will need to integrate the joint density function $p(x,y)$ over the region defined by the constraint. It is very convenient to draw the region ''R'' and the new region define by the constraint $x\geq y$

I will attach a drawing with the region ''R'' and the new region where we need to apply the integral.

If we integrate outside ''R'', given that $p(x,y)=0$ outside ''R'', the integral will be equal to 0 (because of the joint density function).

Inside the rectangle ''R'' and given the constraint $x\geq y$ , we define two new regions : the green region (I) and the blue region (II).

The final step is to integrate in (I) and in (II) and sum ⇒

$\int\int p(x,y) dx dy$ ⇒

$\int \int\limits_1 {p(x,y)} \, dx dy + \int \int\limits_2{p(x,y)} \, dx dy$ , where ''1'' is the green region and ''2'' is the blue region.

⇒ $\int\limits^2_0 \int\limits^x_0 (\frac{1}{16} xy) dy dx$ + $\int\limits^4_2\int\limits^2_0 (\frac{1}{16}xy) dydx$ = $\frac{1}{8}+\frac{3}{4}=\frac{7}{8}=0.875$

We find that $\frac{7}{8}$ of the population satisfy the constraint $x\geq y$ .

6 0

3 years ago