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Gemiola [76]
3 years ago
12

Solve for x. −8x+14≥60 OR −4x+50<58

Mathematics
1 answer:
Reika [66]3 years ago
6 0

Answer: The solution is,

x\leq -\frac{23}{4} or x>-2

Step-by-step explanation:

Given compound inequality,

-8x + 14 ≥ 60 or -4x + 50 < 58,

By the subtraction property of inequality,

-8x + 14 - 14 ≥ 60 - 14 or -4x + 50 - 50  < 58 -50,

-8x ≥ 46 or -4x < 8

By the division property of inequality,

-x \geq \frac{46}{8} or -x < \frac{8}{4}

Using property a > b ⇒ - a < -b,

x\leq -\frac{23}{4} or x>-2

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Find the GCF 4x^5y^3+8x^4y^4+4x^3y^2
Alekssandra [29.7K]

4x^5y^3+8x^4y^4+4x^3y^2

GCF is the greatest common factor

We take GCF of numbers  first and then we take GCF for variables

GCF of 4, 8, 4 is 4 because all the numbers are divisible by 4

For variable, lowest exponent is our GCF

GCF of x^5+x^4+x^3 is x^3

GCF of y^3+y^4+y^2 is y^2

So GCF is 4x^3y^2








6 0
3 years ago
Hey lol i don't really need help on this cuz i could honestly look it up in the notes, but im lazy.
sertanlavr [38]
Theoretically =theory is potentially done, experimental is experiment XD XD. They both have to do with probability
3 0
2 years ago
4 students equally share 2/3 of a pizza
oee [108]

\frac{2}{3} *4

=\frac{2*4}{3}

=\frac{8}{3}

2.67 or 2 \frac{2}{3}

6 0
3 years ago
Emmanuel carves a cone-shaped hollow out of a solid clay cylinder. What is the approximate volume of the solid portion of the cy
bixtya [17]

Answer:

3.14r^2(h-\frac{1}{3}h_1)

Step-by-step explanation:

Let h be the cylinders height and r the radius.

-The volume of a cylinder is calculated as:

V=\pi r^2h

-Since the cone is within the cylinder, it has the same radius as the cylinder.

-Let h_1be the height of the cone.

-The area of a cone is calculated as;

V=\pi r^2 \frac{h}{3}\\\\=\frac{1}{3}\pi r^2h_1

The volume of the  solid section of the cylinder is calculated by subtracting the cone's volume from the cylinders:

V=V_{cy}-V_{co}\\\\=\pi r^2h-\frac{1}{3}\pi r^2 h_1, \pi=3.14\\\\=3.14r^2(h-\frac{1}{3}h_1)

Hence, the approximate area of the solid portion is 3.14r^2(h-\frac{1}{3}h_1)

5 0
3 years ago
Help Asap plz!!!! Thx Give 2<br>0 points
Arisa [49]

Answer:

(1,2) is a solution to both equations

Step-by-step explanation:

To determine if (1,2) is a solution to both equations, substitute into the equation and see if it is true

2x+y =4

2(1) + 2 =4

2+2 =4

4=4

true


y =3x-1

2 = 3(1) -1

2 =3-1

2=2

true

Since both statements are true

(1,2) is a solution to both equations

4 0
3 years ago
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