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SSSSS [86.1K]
3 years ago
12

I need these correct answers nowww

Mathematics
2 answers:
Brums [2.3K]3 years ago
5 0

Step-by-step explanation:

on the picture

I hope it's helpful ❤❤❤

THANK YOU.

faust18 [17]3 years ago
3 0

Answer:

1 - 3/14

2 - 10 5/8

3 - 3/8

4 - 12 1/4

Step-by-step explanation:

numbers 1 and 3 are simple, just multiply the numerators and denominators together.

Numbers 2 and 4, you convert the mixed number to an improper fraction and multiply. Don't forget to switch the answer back to a mixed number if needed!

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1) m − 3 ≥ 3 <br> A) m ≥ −6 : <br> B) m ≥ 9 : <br> C) m ≥ −6 : <br> D) m ≥ 6 :
FromTheMoon [43]

Answer:

The answer should be A and C because they are the same question

Step-by-step explanation:

I just looked at the equations and solved it from there

3 0
3 years ago
Please help! I need help on my math grade problem. Please explain if you could
Rus_ich [418]

Answer:

Answer = 3

Step-by-step explanation:

Constant of proportionality = slope

Slope=\frac{y^{2}-y^{1}  }{x^{2}-x^{1}  } =\frac{12-9}{4-3}=\frac{3}{1}=3

4 0
2 years ago
Read 2 more answers
A sign along the highway says 6% grade for the next 7 mi. How many feet of vertical change there are along those 7 miles. (5280
Tasya [4]

Answer:

Step-by-step explanation:

The length of the road is 7 * 5280 = 36960 feet

The grade is 6% which means that for every 100 feet horizontally, the road rises 6 feet.

6/100 * 36960 = 221760/100 = 2217.6 is the rise.

6 0
2 years ago
How much normally is retaking a high school class?
leonid [27]

Answer:

What do you mean?

Step-by-step explanation:

5 0
2 years ago
Solve the initial-value problem using the method of undetermined coefficients. y"+16y= e^x + x^3.
il63 [147K]
The homogeneous part of the ODE has characteristic equation

r^2+16=0

with roots at r=\pm4i, so the characteristic solution would be

y_c=C_1\cos4x+C_2\sin4x

As a guess for the solution to the nonhomogeneous part, we can try

y_p=ae^x+b_0+b_1x+b_2x^2+b_3x^3
\implies{y_p}''=ae^x+2b_2+6b_3x

Substituting into the ODE gives

(ae^x+2b_2+6b_3x)+16(ae^x+b_0+b_1x+b_2x^2+b_3x^3)=e^x+x^3
17ae^x+(16b_0+2b_2)+(16b_1+6b_3)x+16b_2x^2+16b_3x^3=e^x+x^3
\implies\begin{cases}17a=1\\16b_0+2b_2=0\\16b_1+6b_3=0\\16b_2=0\\16b_3=1\end{cases}\implies a=\dfrac1{17},b_0=0,b_1=-\dfrac3{128},b_2=0,b_3=\dfrac1{16}

so that the particular solution is

y_p=\dfrac1{17}e^x+\dfrac1{16}x^3-\dfrac3{128}x

and the general solution to the ODE is

y=y_c+y_p
y=C_1\cos4x+C_2\sin4x+\dfrac1{17}e^x+\dfrac1{16}x^3-\dfrac3{128}x

Given the initial values, we have

y(0)=4\implies4=C_1+\dfrac1{17}\implies C_1=\dfrac{67}{17}
y'(0)=0\implies0=4C_2+\dfrac1{17}-\dfrac3{128}\implies C_2=-\dfrac{77}{8704}

so that the solution to the IVP is

y=\dfrac{67}{17}\cos4x-\dfrac{77}{8704}\sin4x+\dfrac1{17}e^x+\dfrac1{16}x^3-\dfrac3{128}x
6 0
3 years ago
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