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3 years ago

If James measures 5‘4“ tall how tall is James in inches

Mathematics

1 answer:

Kitty [74]3 years ago

8 0

Answer:

64 inches

Step-by-step explanation:

5 * 12 = 60

60 + 4 = 64

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Four cookies and three cupcakes code $13.25. Five cookies and two cupcakes cost $11.75. Give the system of equations that could

Kaylis [27]

Answer:

cookie: $1.25
cupcake: $2.75

Step-by-step explanation:

Let x and y represent the cost of a cookie and a cupcake, respectively.

4x +3y = 13.25

5x +2y = 11.75 . . . . . . the system of equations

___

By Cramer's rule:

x = (3(11.75) -2(13.25))/(3(5) -2(4)) = 8.75/7 = 1.25

y = (13.25(5) -11.75(4))/7 = 19.25/7 = 2.75

The cost of one cookie is $1.25; the cost of one cupcake is $2.75.

_____

Cramer's rule gives the solution to the system of equations ...

ax +by =c

dx +ey = f

as ...

∆ = bd -ea

x = (bf -ec)/∆

y = (cd -fa)/∆

7 0

3 years ago

4/7m = 2/7( 2m+1 )<br> What is M?<br> Check Solution

aev [14]

4/7m = 2/7(2m + 1)
4/7m = 4/7m + 2/7
4/7m - 4/7m = 2/7
0 = 2/7 (incorrect)

no solution

5 0

3 years ago

Read 2 more answers

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$