Social Security is supplemental retirement income and should not be the only thing a retiree relies on.

Answers

Answer 1

Answer: if its true or false the answer is true

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the length of a rectangle is 50 feet less than its width. If the perimeter of the field is 840 feet, find the length of the field.

Answers

$look\ at\ the\ picture\n\n2a+2(a-50)=840\n\n2a+2a-100=840\n\n4a-100=840\ \ \ /+100\n\n4a=940\ \ \ \ /:4\n\na=235\ (ft)\n\na-50=235-50=185\ (ft)\n\nAnswer:lenght=185\ ft;\ width=235\ ft.$

Jose asks his friends to guess the higher of two grades he received on his math tests. He gives them two hints. The difference of the two grades is 16. The sum of one-eighth of the higher grade and one-half of the lower grade is 52. The system that represents his scores is below.. . x – y = 16. . x + y = 52. . What is the higher grade of Jose’s two tests?. . 48.
52.
80.
96

Answers

x - y = 16
1/8 x + 1/2 y = 52 / * 2 ( multiply both sides by 2 )
-------------------------
x - y = 16
+
1/4 x + y = 104
----------------------
1.25 x = 120
x = 120 : 1.25
x = 96
Answer: D ) 96

your answer sould be the last option (D) 96

The regular price of a TV is $475. It's currently on sale for 20% off. What is the discount if you bought the TV during the sale?

Answers

Answer:

475 x 0.20 = $95

475 - 95 = $380

Step-by-step explanation:

so your answer is $380!

hope this helps! :)

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Answers

The area bounded by the functions f(x) and g(x) in graph below.

The given function are f(x)=x² and g(x)=√x.

What is the function?

Functions are the fundamental part of the calculus in mathematics. The functions are the special types of relations. A function in math is visualized as a rule, which gives a unique output for every input x.

To find the area between two curves defined by functions, integrate the difference of the functions. If the graphs of the functions cross, or if the region is complex, use the absolutevalue of the difference of the functions.

Area bounded = |x²-√x|

Find the domain by finding where the expression is defined.

Interval Notation:

[0,∞)

Set-Builder Notation:{x|x≥0}

Therefore, the area bounded by the functions f(x) and g(x) in graph below.

To learn more about the function visit:

brainly.com/question/28303908.

#SPJ2

Answer:

$\large\boxed{1(1)/(3)\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=√(x)\qquad\text{square of both sides}\n\n(x^2)^2=\left(√(x)\right)^2\n\nx^4=x\qquad\text{subtract}\ x\ \text{from both sides}\n\nx^4-x=0\qquad\text{distribute}\n\nx(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\n\nx^3-1=0\qquad\text{add 1 to both sides}\n\nx^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_(0)^1(√(x))dx-\int\limits_(0)^1(x^2)dx=(*)\n\n\int(√(x))dx=\int\left(x^(1)/(2)\right)dx=(2)/(3)x^(3)/(2)=(2x√(x))/(3)\n\n\int(x^2)dx=(1)/(3)x^3\n\n(*)=\left((2x√(x))/(2)\right]^1_0-\left((1)/(3)x^3\right]^1_0=(2(1)√(1))/(2)-(2(0)√(0))/(2)-\left((1)/(3)(1)^3-(1)/(3)(0)^3\right)\n\n=(2(1)(1))/(2)-(2(0)(0))/(2)-(1)/(3)(1)}+(1)/(3)(0)=2-0-(1)/(3)+0=1(1)/(3)$