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EleoNora [17]
3 years ago
13

If B = y + 2 and C Y-7, find an expression that equals B + 2C in standard form.

Mathematics
2 answers:
Lerok [7]3 years ago
7 0

Answer:

3y-12

Step-by-step explanation:

b=y+2, and c=y-7. We know that b=y+2, so b+2c can be written as y+2+2c. We also know c=y-7, so we can write it as y+2+2y-14, so we have 3y-12.

statuscvo [17]3 years ago
3 0

Answer:

I think the answer is 5+ 6c

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Naddika [18.5K]
Estimate 3458092=3500000~; 57=60~.
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Solve 2x^2+16x+34=0?????
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2(x^{2}+8x+17)

Step-by-step explanation:

Factor out 2 of the equation.

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Step-by-step explanation:

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"You are dating Moon rocks based on their proportions of uranium-238 (half-life of about 4.5 billion years) and its ultimate dec
Olenka [21]

Step-by-step explanation:

a.

Initial mass of the isotope = x

Time taken by the sample to decay its mass by 41% = t

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

where,

N_o = initial mass of isotope  = x

N = mass of the parent isotope left after the time, (t)  = 59% of x = 0.59x

t_{\frac{1}{2}} = half life of the isotope  = 4.5 billion years

\lambda = rate constant

Now put all the given values in this formula, we get

0.59x=x\times e^{-(\frac{0.693}{\text{4.5 billion years}})\times t}

t = 3.4 billion years

The age a rock is 3.4 billion years.

b.

Initial mass of the isotope = x

Time taken by the sample to decay its mass by 35%= t

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

where,

N_o = initial mass of isotope  = x

N = mass of the parent isotope left after the time, (t)  = 65% of x = 0.65x

t_{\frac{1}{2}} = half life of the isotope  = 4.5 billion years

\lambda = rate constant

Now put all the given values in this formula, we get

0.65x=x\times e^{-(\frac{0.693}{\text{4.5 billion years}})\times t}

t = 2.8 billion years

The age a rock is 2.8 billion years.

5 0
4 years ago
A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5.5 . What are the dimensions o
cluponka [151]

Answer:

r = 3.667

h = 1.5

Step-by-step explanation:

Given:-

- The base radius of the right circular cone, R = 5.5

- The height of the right circular cone, H = 4.5

Solution:-

- We will first define two variables that identifies the volume of a cylinder as follows:

                                r: The radius of the cylinder

                                h: The height of cylinder

- Now we will write out the volume of the cylinder ( V ) as follows:

                                V = \pi*r^2h

- We see that the volume of the cylinder ( V ) is a function of two variables ( don't know yet ) - ( r,h ). This is called a multi-variable function. However, some multi-variable functions can be reduced to explicit function of single variable.

- To convert a multi-variable function into a single variable function we need a relationship between the two variables ( r and h ).

- Inscribing, a cylinder in the right circular cone. We will denote 5 points.

              Point A: The top vertex of the cone

              Point B: The right end of the circular base ( projected triangle )

              Point C: The center of both cylinder and base of cone.

              Point D: The top-right intersection point of cone and cylinder

              Point E: Denote the height of the cylinder on the axis of symmetry of both cylinder and cone.  

- Now, we will look at a large triangle ( ABC ) and smaller triangle ( ADE ). We see that these two triangles are "similar". Therefore, we can apply the properties of similar triangles as follows:

                              \frac{AC}{AE} = \frac{BC}{DE}  \\\\\frac{H}{H-h} = \frac{R}{r}

- Now we can choose either variable variable to be expressed in terms of the other one. We will express the height of cylinder ( h ) in term of radius of cylinder ( r ) as follows:

                             H- h = r\frac{H}{R} \\\\h = \frac{H}{R}*(R-r)

- We will use the above derived relationship and substitute into the formula given above:

                            V = \pi r^2 [ \frac{H}{R}*(R - r )]\\\\V = \frac{\pi H}{R}.r^2.(R-r)

- Now our function of volume ( V ) is a single variable function. To maximize the volume of the cylinder we need to determine the critical points of the function as follows:

                            \frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-2r^2 - r^2 )\\\\\frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-3r^2 ) = 0\\\\(2rR-3r^2 ) = 0\\\\2R -3r = 0\\\\r = \frac{2}{3}*R

- We found the limiting value of the function. The cylinder volume maximizes when the radius ( r ) is two-thirds of the radius of the right circular cone.

- We can use the relationship between the ( r ) and ( h ) to determine the limiting value of height of cylinder as follows:

                          h = \frac{H}{R} * ( R - \frac{2}{3}R)\\\\h = \frac{H}{3}

- The dimension of the inscribed cylinder with maximum volume are as follows:

                         r = \frac{2}{3}*5.5 = 3.667\\\\h = \frac{4.5}{3} = 1.5

Note: When we solved for the critical value of radius ( r ). We actually had two values: r = 0 , r = 2R/3. Where, r = 0 minimizes the volume and r = 2R/3 maximizes. Since the function is straightforward, we will not test for the nature of critical point ( second derivative test ).

7 0
4 years ago
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