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Advocard [28]
3 years ago
8

Complete the square to rewrite y=x(squared)-6x+14 in vertex form.

Mathematics
2 answers:
WITCHER [35]3 years ago
4 0
Y = x² - 6x + 14
y = x² - 6x + 3² - 3² -14
y = (x - 3)² - 9 - 14
y = (x - 3)² - 23

------------------------------------------------
Answer : y = (x - 3)² - 23
------------------------------------------------
dem82 [27]3 years ago
3 0
Y
=
x
2
+
6
x
−
14 this is the answer
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Bill weighs 120 lbs and is gaining ten lbs each month. Phil weighs 150 lbs and is gaining four lbs each month. How many months w
vaieri [72.5K]
First, let's write two expressions, letting x= the number of months which have elapsed:

Bill: 120 + 10x
Phil: 150 + 4x

If we set them equal to each other, then solve for x, that will be the number of months where their weights equal each other:

120+10x = 150 + 4x   [starting equation]
-120  -4x    -120 -4x    [ subtract 120 from both sides, and 4x from both sides, to isolate the term with the variable]

<u>6x</u> = <u>30</u>     [divide both sides by 5]
 5      5

x=6. They will weigh the same in six months.

5 0
3 years ago
Read 2 more answers
2y)(2y + 8) = ? A. 4xy – 16x – 4y2 – 16y B. 4xy + 16x – 4y2 – 16y C. 4xy – 4y2 D. 4xy + 16x + 4y2 + 16y 8…
Vilka [71]
4y^2 + 16y
You distribute the 2y into the 2y+8
8 0
3 years ago
Cesar is excited that he only has 12 months left before he pays off his credit card completely. His current balance is $3,750 an
kodGreya [7K]
The answer is.

b.

$91.44
6 0
3 years ago
Which answer is equivalent to 4 1/3*4 1/6?<br> a. 16<br> b .6√​16<br> c. 2<br> d. 3√​4
Talja [164]

Answer:

<h2>c. 2</h2>

Step-by-step explanation:

4^\frac{1}{3}\cdot4^\frac{1}{6}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=4^{\frac{1}{3}+\frac{1}{6}}=4^{\frac{1\cdot2}{3\cdot2}+\frac{1}{6}}=4^{\frac{2}{6}+\frac{1}{6}}=4^\frac{2+1}{6}=4^\frac{3}{6}=4^\frac{3:3}{6:3}=4^{\frac{1}{2}}\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\=\sqrt[2]{4^1}=\sqrt4=2

5 0
3 years ago
A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
3 0
3 years ago
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