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lyudmila [28]
4 years ago
5

3x - 4 < 8 and 2x+ 2 > 4 20 points crq

Mathematics
1 answer:
Gekata [30.6K]4 years ago
4 0

Answer:

x<4 and x>1

Step-by-step explanation:

1) 3x-4<8

Add 4 to both sides first to isolate x.

3x-4<8

  <u>+4 +4</u>

3x<12

Divide both sides by 3.

3x<12

<u>/3  /3</u>

x<4

2) 2x+2>4

Subtract both sides by 2 to isolate x.

2x+2>4

   <u>-2  -2</u>

2x>2

Divide both sides by 2.

2x>2

<u>/2  /2</u>

x>1

Hope this helped!! Have an amazing day :3

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Find the sum of all integers 30 to 80, INCLUSIVE. (Is there a formula to do this?)
Marina86 [1]
Yes there is actually a formula for this. This is an arithmetic progression with a common difference of 1. 

The sum of terms can be solved by using the formula

S = (n/2) * (a1 + an)

where S is the sum of the terms
n is the number of terms 
a1 is the first term
an is the last term

we can calculate the number of terms, n by using the formula

an = a1 + (n-1)*d

d is the common difference equal to 1 

therefore n = 51

substituting to the first equation

S = 2805
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3 years ago
Mario started his homework at 3:30. He finishes 25 minutes later. What time did Mario finish his homework
jenyasd209 [6]
He finishes his homework at 3:55
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3 years ago
A rectangular gate is made using 6 straight pieces of steel.
Sunny_sXe [5.5K]

Answer:

Total weight is 150 kg

Step-by-step explanation:

Firstly, we need the weight of the diagonal

To get this, we shall use the Pythagoras’ theorem which states that the square of the hypotenuse equals the sum of the squares of the two other sides

The diagonals stand for the hypotenuse

Thus, we have that;

d^2 = 5^2 + 12^2

d^2 = 25 + 144

d^2 = 169

d^2 = 13^2

d = 13

The 6 sides technically includes the whole rectangle sides, then the two diagonals

The total length here is;

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Weight of 60 m will be ;

60 * 2.5 = 150 kg

4 0
3 years ago
If d=8, find the value of 3d. _____
yan [13]

Answer:

24

Step-by-step explanation:

3d

= 3 × 8

= 24

Hope it helps you:)

3 0
2 years ago
Read 2 more answers
CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS
finlep [7]

Due to the symmetry of the paraboloid about the <em>z</em>-axis, you can treat this is a surface of revolution. Consider the curve y=x^2, with 1\le x\le2, and revolve it about the <em>y</em>-axis. The area of the resulting surface is then

\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6

But perhaps you'd like the surface integral treatment. Parameterize the surface by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k

with 1\le u\le2 and 0\le v\le2\pi, where the third component follows from

z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2

Take the normal vector to the surface to be

\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}

Then the area of the surface is

\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

which reduces to the integral used in the surface-of-revolution setup.

7 0
3 years ago
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