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PolarNik [594]
2 years ago
8

PLEASE HELP ME I BEG IMMA FAILLLLLLLLLLLLLLLL

Mathematics
1 answer:
Irina18 [472]2 years ago
4 0

Answer:

7:1

Step-by-step explanation:

There are two cubes, one with 14 cm and one with 2 cm. 14:2 is equal to 7:1.

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The gasoline prices in seven stayed are $1.96,$2.09,$1.79,$1.61,$1.75,$2.11 and $1.84 what is the median gasoline price
Digiron [165]

The median is the one in the middle. Sorting the data gives

$1.61, $1.75, $1.79, $1.84, $1.96, $2.09, $2.11

Seven data points and the fourth one is 1.84.

Answer: $1.84


3 0
3 years ago
Equivalent expression to 3(4m-2)-2(m+5
wolverine [178]

Answer:

10m - 16

Step-by-step explanation:

1.  Perform the indicated multiplication.  We get 12m - 6 - 2m - 10.

2.  Combine like terms:  We get 10m - 16

4 0
3 years ago
Can someone help me plead
andrew-mc [135]

Answer:

a. 2^3

b. 3^4

c. 4^3 × 5^2

d. 9^4 × 7^2

Step-by-step explanation:

The following equations are given

a. 2 × 2× 2

b. 3 × 3 × 3 × 3

c. 4 × 4 × 4 × 5 × 5

d. 9 × 7 × 9 × 9 × 7 × 9

We need to find the index notation for the above equations

a. 2^3

b. 3^4

c. 4^3 × 5^2

d. 9^4 × 7^2

In this way it should be done

The same would be relevant

4 0
2 years ago
What is the formula to find the volume of the solid ?
tester [92]

Answer:

Step-by-step explanation:

This solid figure is a cone,

Volume=(πr²H)/3 cubic units

r ----> radius

H -------> height

6 0
3 years ago
Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.
Mama L [17]

Multiplying both sides by y^2 gives

xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1

so that substituting v=y^3 and hence \frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx} gives the linear ODE,

\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1

Now multiply both sides by 3x^2 to get

x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2

so that the left side condenses into the derivative of a product.

\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2

Integrate both sides, then solve for v, then for y:

x^3v=\displaystyle\int3x^2\,\mathrm dx

x^3v=x^3+C

v=1+\dfrac C{x^3}

y^3=1+\dfrac C{x^3}

\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}

6 0
3 years ago
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