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otez555 [7]
2 years ago
9

Taylor has a $30 gift card that she can spend at the store. She has already bought a $9 picture frame, and she wants to buy $3 j

ournals with the leftover money on the card. How many journals can she buy without going over the card's limit? What is the variable
Mathematics
1 answer:
Lorico [155]2 years ago
7 0

Answer:

9

Step-by-step explanation:

total is 30

spent is $9

this is total money minus the amount of the picture frame

30-9

21 divide by 3

the total sum of journals bought was 9

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Solve the following inequality: 38 < 4x + 3 + 7 – 3x.
DochEvi [55]
38 < 4x + 3 +7 - 3x (original equation)
38 < 4x - 3x + 3 + 7 (combine like terms)
38 < x + 10 (simplify)
38 - 10 < x + 10 - 10 (subtract 10 from both sides to get (x) alone)
28 < x (simplify)
x > 28 (switch to get x on the left (its proper equation writing >.<) ) 
7 0
2 years ago
Read 2 more answers
Each locker is shaped like a rectangular prism. Which has more storage space? explain/
seropon [69]
Locker 1 has more storage space because it has a greater volume.
The volume of locker 1 is 8,640 (48x15x12)
The volume of locker 2 is 7,200 (60x10x12)

I hope this helps.
4 0
2 years ago
Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common ch
enot [183]

Answer:

y=\frac{-7t^2+22t-7}{7t-22}

Step-by-step explanation:

We are given that

Initial value problem

y'=(t+y)^2-1, y(3)=4

Substitute the value z=t+y

When t=3 and y=4 then

z=3+4=7

y'=z^2-1

Differentiate z w.r.t t

Then, we get

\frac{dz}{dt}=1+y'

z'=1+z^2-1=z^2

z^{-2}dz=dt

Integrate on both sides

-\frac{1}{z}dz=t+C

z=-\frac{1}{t+C}

Substitute t=3 and z=7

Then, we get

7=-\frac{1}{3+C}

21+7C=-1

7C=-1-21=-22

C=-\frac{22}{7}

Substitute the value of C then we get

z=-\frac{1}{t-\frac{22}{7}}

z=\frac{-7}{7t-22}

y=z-t

y=\frac{-7}{7t-22}-t

y=\frac{-7-7t^2+22t}{7t-22}

y=\frac{-7t^2+22t-7}{7t-22}

8 0
2 years ago
Divide 60 kg in the ratio of 5 : 7.
CaHeK987 [17]
The ratio can be written in parts as shown below;
\frac{5}{12} : \frac{7}{12}
Then, 60 can be divided as shown below;
\frac{5}{12} *60: \frac{7}{12} *60
25kg:35kg
6 0
2 years ago
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L = loudness, in decibels (dB); I = sound intensity, in watts/m2; I0 = 10−12 watts/m2
irina [24]
The sound intensity of the Pile Driver is 39.5
 or nearly 40 times the sound intensity of the jackhammer.


Given with Loudness in dB for pile driver = 112 dB
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First,
112dB/10 = 11.2 

Then we'll use this as exponent of 10
(10)^(11.2) = 1.5849 * 10 ^ 11

Then use the equation of Watts per square meter to find the intensity:
I / (10^-12 W/m^2) =1.5849 * 10 ^ 11
I = sound intensity = 0.158

Then compare:

Sound intensity of Pile Driver/ Sound intensity of Jackhammer
(0.158) / (0.004)
= 39.5
or nearly 40 times the jackhammer.
3 0
2 years ago
Read 2 more answers
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