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scZoUnD [109]
3 years ago
6

2- Describe the cross section

Mathematics
1 answer:
kicyunya [14]3 years ago
4 0

The cross-section cut by the plane yields a quadrilateral with length and width equal to the length and width of the face that is parallel to the plane.

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87 24/25 as a decimal
adoni [48]
87.96 just divide 24 by 25 and keep the whole number as is
8 0
3 years ago
Read 2 more answers
Please procedure and solution <br><br>2 x |-6|-3 x |2|+|-14|=​
lora16 [44]

Applying the rule of modulus for given values, the expression will result to 20.

<h3>What is Modulus or Absolute Values</h3>

The modulus of a value for example |a|=a if a is greater than or equal to zero

and also

|a|=-a if a is less than zero

In the expression given; the modulus of -6 written as

|-6|=-(-6) this is because the value is less than zero

The modulus of 2 written as; |2|=2 this is because the value is greater than zero. and;

|The modulus of -14 written as; |-14|= -(-14) since the value is less than zero.

Hence, we can rewrite the expression as;

we deal with multiplication first following the rules of BODMAS

2×-(-6)-3×2+[-(-14)]

2×6-6+14

and thus adding values

12+14-6

and finally we subtract;

26-6

which will results to the solution of 20.

5 0
2 years ago
Your drawer contains 10 red socks and 7 blue socks. It's to dark to see which are which,but you grab two anyway.What is the prob
den301095 [7]
P(both red)=(10/17)(9/16)
P(both blue)=(7/17)(6/16)

P(both same)= sum of above 2 = (90+42)/(16*17)=132/(16*17)=<span>0.485

coz both same means  both blue or both red
</span>
4 0
3 years ago
Read 2 more answers
Lim x-1 x2 - 1/ sin(x-2)
balu736 [363]

Answer:

           \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=0

Explanation:

Assuming the correct expression is to find the following limit:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}

Use the property the limit of the quotient is the quotient of the limits:

         \lim_{x \to 1}\frac{x^2-1}{sin(x-2)}=\frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}

Evaluate the numerator:

          \frac{\lim_{x \to 1}x^2-1}{\lim_{x \to 1}sin(x-2)}=\frac{1^2-1}{\lim_{x \to1}sin(x-2)}=\frac{0}{\lim_{x \to 1}sin(x-2}

Evaluate the denominator:

  • Since         \lim_{x \to1}sin(x-2)\neq 0

                  \frac{0}{\lim_{x \to1}sin(x-2)}=0

4 0
3 years ago
How many x-intercepts would the function shown below have? <br> F(x) =x^2 (x^2+9) (x^2-3) (5x+1) ^2
-BARSIC- [3]

Step-by-step explanation:

step 1. x intercepts are where y = 0

step 2. y = 0 when factors are equal to zero

step 3. the x intercepts are 0, sqr(3), -sqrt(3), -1/5

step 4. answer: 4.

6 0
2 years ago
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