1. Jack is 30 years old and Alice is 35 years old. What percent of Alice's age is Jack, rounded to the nearest tenth of a percent?

Answers

Answer 1

Answer: the answer would be 57percent

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NEED HELP ASAP!!!!What is the area of this composite shape?

Enter your answer in the box.

in²

Image shows a polygon that looks like a rectangle with a triangle attach to the left side of it closer to the top. The polygon has 5 sides. The top side of the polygon is 12 inches. The right side is perpendicular to the top side and is 7 inches. The bottom side is perpendicular to the right side and is 8 inches. The left side has two segment parts. The bottom part of the left side is perpendicular to the bottom and is 4 inches high. The last segment is slanted connecting the 4 inch side to the top side. This side is not labeled

Answers

Answer:

62 square inches

Step-by-step explanation:

we know that

The area of the composite figure is equal to the area of a rectangle plus the area of a triangle

$A=(8)(7)+(1)/(2)(7-4)(12-8)$

$A=56+6=62\ in^2$

If r = 10 and s = 31, find R. Round to the nearest tenth

Answers

Answer: c 17.9°

Step-by-step explanation:

Answer:

the answer is c

Step-by-step explanation:

I took the test

If sine of x equals square root of 2 over 2, what is cos(x) and tan(x)? Explain your steps in complete sentences.

Answers

sin $x$ = $(opposite)/(hypotenuse)$ = $( √(2) )/(2)$
opposite= $√(2)$
hypotenuse=2

$Adjacent^(2)$ + $(√(2 ))^(2 )$ = $2 ^(2)$
$Adjacent^(2)$ +2=4
$Adjacent^(2)$ =2
Adjacent= $√(2)$

cos $x$ = $(ajc)/(hyp)= ( √(2) )/(2)$
Tan $x$ = $(opp)/(adj) = ( √(2) )/( √(2) )$

Could you please solve me this exercice: :3 $x^2-5x+√(x^2-5x+7)=5$

Thank you a lot! :3 :3 :3

Answers

$Domain:x^2-5x+7\geq0\n\na=1;\ b=-5;\ c=7\n\n\Delta=b^2-4ac\to\Delta=(-5)^2-4\cdot1\cdot7=25-28=-3 < 0\n\ntherefore\ D:x\in\mathbb{R}$

$Find\ minimum\ value\ of\ x^2-5x+7\n\ny_(min)=(-\Delta)/(4a)\to y_(min)=(-(-3))/(4\cdot1)=\boxed{(3)/(4)}\n-----------------------------\nx^2-5x+√(x^2-5x+7)=5\ \ \ \ \ |add\ 7\ to\ both\ sides\n\nx^2-5x+7+√(x^2-5x+7)=12\n\nsubtitute\ t=x^2-5x+7\ (t\geq(3)/(4))\n\nt+√(t)=12\ \ \ \ |subtract\ t\ from\ both\ sides\n\n√(t)=12-t\ \ \ \ \ \ |square\ both\ sides$

$t=(12-t)^2\n\nt=12^2-2\cdot12\cdot t+t^2\n\nt=144-24t+t^2\n\nt^2-24t+144=t\ \ \ \ \ |subtract\ t\ from\ both\ sides\n\nt^2-25t+144=0\n\nt^2-9t-16t+144=0\n\nt(t-9)-16(t-9)=0\n\n(t-9)(t-16)=0\iff t-9=0\ or\ t-16=0\n\nt=9\ or\ t=16$

$t+√(t)=12\ therefore\ t=16\ is\ not\ a\ solution.$

$t=9\to x^2-5x+7=9\ \ \ \ |subtract\ 9\ from\ both\ sides\n\nx^2-5x-2=0\n\na=1;\ b=-5;\ c=-2\n\n\Delta=(-5)^2-4\cdot1\cdot(-2)=25+8=33\n\nx=(-b\pm\sqrt\Delta)/(2a)\to x=(5\pm√(33))/(2)$

$Answer:\n\boxed{x=(5-√(33))/(2)\ or\ x=(5+√(33))/(2)}$

if xy=0 then x or/and y must be 0 so we try to get it to zero later

so first we get the square root on one side by itself
so subtract x^2-5 from both sides (-x^2+5 to both sides)

√(x^2-5x+7)=-x^2+5x+5
square both sides (or both sides to the ^2 power)
gets rid of the √
x^2-5x+7=(-x^2+5x+5)^2=x^4-10x^3+15x^2+50x+25
so subtract x^2 from both sides
-5x+7=x^4-10x^3+14x^2+50x+25
add 5x to both sides
7=x^4-10x^3+14x^2+55x+25
subtract 7 from both sides
0=x^4-10x^3+14x^2+55x+18
factor
(x^2-5x-9)(x^2-5x-2)=0
(x^2-5x-9)=0
(x^2-5x-2)=0
I'm not sure how to proceed but if you can find the factors of (x^2-5x-2) and (x^2-5x-9) and set them to zero, you can get the answers

after 1 second the ball is 76 feet in the air: after 2 seconds it is 144 feet in the air. Find the height of the function. h(x)= ?

Answers

General equation for height, h
h = ut + 0.5gt^2

at t=1s, h = 76 feet
76 = u*1 + 0.5*a*1^2
76 = u + 0.5a

at t=2s, h = 144 feet
144 = u*2 + 0.5*a*2^2
144 = 2u + 2a
72 = u + a

Subtract 2 equations
0.5a = -4
a = -8
u = 80

So the equation,
h(x) = 80t + 0.5*(-8) t^2
h(x) = 80t - 4t^2

Whic of these is an irrational number A.0 B1.9 C.? D.1/2