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Mariana [72]
3 years ago
5

Plz help with Part A

Mathematics
1 answer:
aksik [14]3 years ago
4 0
4/5= 4 x 2=8
5 x 2=10

5/8=5 x 3 =15
6 x 3= 18
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The volume of the shipping box needs to be 1,144 cubic inches. The equation that models the volume of the shipping box is 8(n +
Snezhnost [94]

Answer:

see below

Step-by-step explanation:

8(n + 2)(n + 4) = 1,144

FOIL

8(n^2 +2n+4n+8) = 1144

Divide each side by 8

8/8(n^2 +2n+4n+8) = 1144/8

(n^2 +2n+4n+8) = 143

Combine like terms

n^2 +6n+8 = 143

Subtract 143 from each side

n^2 +6n+8 -143= 0

Combine like terms

n^2 +6n -135 =0

Factor

What two terms multiply to -135 and add to 6

-9*15 =-135

-9+15 = 6

(n-9) (n+15) =0

Using the zero product property

n-9 =0    n+15=0

n = 9   n=-15

The length cannot be negative so n = -15 cannot be  a solution

n =9

7 0
3 years ago
Read 2 more answers
Under a dilation, the point (3,5) is moved to (6,10) What is the scale factor of the dilation?
nlexa [21]

The scale factor of the dilation is 2.

Option D is correct.

Step-by-step explanation:

To find the scale factor of dilation, we find ratio of similar sides.

We are given  the point (3,5) is moved to (6,10) under dilation.

Scale Factor = 6/3 = 2

or

Scale Factor = 10/5 = 2

The scale factor of the dilation is 2.

Option D is correct.

Keywords: scale factor of the dilation

Learn more about scale factor of the dilation at:

  • brainly.com/question/2480897
  • brainly.com/question/4706270
  • brainly.com/question/5563823

#learnwithBrainly

5 0
3 years ago
Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth
taurus [48]

Answer:

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\

It is a homogeneous solution:

y_h(t)=c_1e^{-2i t}+c_2e^{2i t}

Now, we finding a particular solution.

y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\

we get

y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

So, solution of  the differential equation is

y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos  2x+c_1e^{-2it}+c_2e^{2it}\\

7 0
3 years ago
Find the equation of the line.
Grace [21]

Answer:

m:-\frac{1}{3}

Step-by-step explanation:

Pick two points (0,5) and (3,4)

use the formula to find slope: \frac{y2-y1}{x2-x1}

Insert values: \frac{4-5}{3-0}

m:-\frac{1}{3}

:-)

7 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
shtirl [24]

Answer:

It would be the third one hope i helped

Step-by-step explanation:

8 0
3 years ago
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