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3 years ago

14

Katie wants to hang a painting in a gallery. The painting and frame must have an area of 45 square feet. The painting is 6 feet

wide by 7 feet long. Which quadratic equation can be used to determine the thickness of the frame, x?

1 answer:

Luda [366]3 years ago

5 0

Each side should increase by 2x (left & right, and up & down) so that to match the condition :
7+2x)(6+2x) = 45
Solving it will give you: 4x² + 26x - 3

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Please Help And NO LINKS OR U BE REPORTED!

Elis [28]

Answer:

I believe your answer is B

4 0

3 years ago

The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0

2 years ago

Read 2 more answers

Suppose that an object moves along the y-axis so that its location is y=x2+3x at time x. (Here y is in meters and x is in second

vfiekz [6]

Answer:

a) 13 m/s

b) (15 + h) m/s

c) 15 m/s

Step-by-step explanation:

if the location is

y=x²+3*x

then the average velocity from 3 to 7 is

Δy/Δx=[y(7)-y(3)]/(7-3)=[7²+3*7- (3²+3*3)]/4= 13 m/s

then the average velocity from x=6 to to x=6+h

Δy/Δx=[y(6+h)-y(6)]/(6+h-6)=[(6+h)²+3*(6+h)- (6²+3*6)]/h= (2*6*h+3*h+h²)/h=2*6+3= (15 + h) m/s

the instantaneous velocity can be found taking the limit of Δy/Δx when h→0. Then

when h→0 , limit Δy/Δx= (15 + h) m/s = 15 m/s

then v= 15 m/s

also can be found taking the derivative of y in x=6

v=dy/dx=2*x+3

for x=6

v=dy/dx=2*6+3 = 12+3=15 m/s

7 0

3 years ago

How does the graph of y = sec(x + 3) – 7 compare with the graph of y = sec(x)?

Alla [95]

Answer:

The function y = sec(x) shifted 3 units left and 7 units down .

Step-by-step explanation:

Given the function: y = sec(x)

If k is any positive real number, then the graph of f(x) - k is the graph of y = f(x) shifted downward k units.

If p is a positive real number, then the graph of f(x+p) is the graph of y=f(x) shifted to the left p units.

The function $y = \sec(x+3)-7$ comes from the base function y= sec(x).

Since 3 is added added on the inside, this is a horizontal shift Left 3 unit, and since 7 is subtracted on the outside, this is a vertical shift down 7 units.

Therefore, the transformation on the given function is shifted 3 units left and 7 units down

3 0

3 years ago

Read 2 more answers

Ok is it 15/28 or 75/135 thatx is evuviulent to 25/45

Aleonysh [2.5K]

You also need to brake down this answer farther.

4 0

3 years ago