dieon has 3/4 pound of dog treats he will give his dog 1/8 of treats everyday how long will the bag last

Answers

Answer 1

Answer: Amount of dog treat that Dieon has = 3/4 pound
Amount of dog treat that Dieon gives his dog each day = 1/8
Let us assume the number of days the dog treat will last = x
Then
(1/8) * x = (3/4)
x/8 = 3/4
x = (3 * 8)/4
= 3 * 2
= 6
So we can see from the above deduction that the dog treat will last for 6 days. I hope the procedure is clear enough for you to understand.

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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

What are the answers to the picture?

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Answer: Brailiest plz?

prt 1 (42°) 48° is complementary

prt2 (42°) 138° is supplementary

prt1 (81°) 9° is complementary

prt2 (81°) 99° is supplementary

Step-by-step explanation:

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Step-by-step explanation:

Solve for x: 5|2x − 2| + 8 = 18.x = 2, x = −2

x = −2, x = 0

x = 2, x = 0

x = 0, x = 1 over 2