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Reptile [31]
3 years ago
5

Help me please with this question

Mathematics
1 answer:
docker41 [41]3 years ago
6 0

Answer:

P=152

A=79

Step-by-step explanation:

add all the sides

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Find the gcf and lcm of 20x^2y^3 60x^3y^2| 40xy^2
LenaWriter [7]
Gcf is the grouping of all the factors common to both groups
lcm is a group of all the factors that has all the elements of both groups

factor
20x^2y^3=2*2*5*x*x*y*y*y
60x^3y^2=2*2*3*5*x*x*x*y*y
40xy^2=2*2*2*5*x*y*y

the GCF=2*2*5*x*y*y=20xy^2
the LCM=2*2*2*3*5*x*x*x*y*y*y=120x^3y^3
7 0
3 years ago
If 3x^2 + y^2 = 7 then evaluate d^2y/dx^2 when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -,
S_A_V [24]
Taking y=y(x) and differentiating both sides with respect to x yields

\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0

Solving for the first derivative, we have

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x}y

Differentiating again gives

\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0

Solving for the second derivative, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

Now, when x=1 and y=2, we have

\dfrac{\mathrm d^2y}{\mathrm dx^2}\bigg|_{x=1,y=2}=-\dfrac{3\cdot2^2+9\cdot1^2}{2^3}=\dfrac{21}8\approx2.63
3 0
3 years ago
(3x-5)/(x-5)&gt;=0 <br> How do I get the answers for this problem?
Diano4ka-milaya [45]
\frac{3x-5}{x-5} > 0 

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: \frac{3x-5}{x-5} = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = \frac{5}{3}
Sixth, from the values of x above, we have these 3 intervals to test: 
x < \frac{5}{3}
\frac{5}{3} < x < 5
x > 5
Seventh, pick a test point for each interval. 

1. For the interval x < \frac{5}{3} :
Let's pick x - 0. Then, \frac{3x0-5}{0-5} > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval \frac{5}{3} < x < 5:
Let's pick x = 2. Then, \frac{3x2-5}{2-5} > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, \frac{3x6-5}{6-5} > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x < \frac{5}{3} and x > 5

Answer: x < \frac{5}{3} and x > 5

3 0
3 years ago
Use the long division method to divide 120÷24​
Natali5045456 [20]

Step-by-step explanation:

120÷24

    

    0   24⟌120

    0   24⟌120   0  

    0   24⟌120   -0      1

    00  24⟌120   -0      12 

   0   24⟌120   -0      12 

    00  24⟌120   -0      12    - 0     12 

   00  24⟌120   -0      12    - 0     120

4 0
3 years ago
Read 2 more answers
What is the difference in length between the shortest and longest worm?<br><br>I don't understand
eduard
Shortest is 3/4 longest is 1 1/2. 1 1/2 -3/4=3/4
8 0
3 years ago
Read 2 more answers
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