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3 years ago

Two distinct planes intersect describe their intersection

2 answers:

7 0

If two planes intersect, then their intersection is a line (Postulate 6). A line contains at least two points (Postulate 1). If two lines intersect, then exactly one plane
contains both lines (Theorem 3).

Sholpan [36]3 years ago

5 0

Answer:

The answer is line.

Step-by-step explanation:

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Morgan uses 1/4 of her supply of raisins to make a trail mix and 3/8 of her supply of raisins to make cookies.If Morgan uses 5 p

Elden [556K]

1/4=25%
3/8=37.5%
37.5%+25%=62.5%
62.5%=0.625
5/0.625=8
Morgan has 8 pounds of raisins in her supply.

3 0

3 years ago

Select the factors that can be divided out to simplify the multiplication problem: 4/26 x 13/20. Select all that apply. A. 2 B.

Tcecarenko [31]

$\dfrac{4}{26} \times \dfrac{13}{20}$

-------------------------------------
Divide by factor of 4.
-------------------------------------
*Or divide by the factor of 2 twice

$\dfrac{1 \times 4}{26} \times \dfrac{13}{5 \times 4}$

$\dfrac{1}{26} \times \dfrac{13}{5}$

-------------------------------------
Divide by factor of 13.
-------------------------------------

$\dfrac{1}{2 \times 13} \times \ \dfrac{1 \times 13}{5}$

$\dfrac{1}{2} \times \dfrac{1 }{5}$

-------------------------------------
Combine into single fraction.
-------------------------------------
$\dfrac{1}{10}$

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Answer: The factors are 2, 4 and 13
--------------------------------------------------------------------------

3 0

4 years ago

Read 2 more answers

Should the object be measured in grams or in kilograms?

Mashcka [7]

Feather: grams
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6 0

3 years ago

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Covert 3/8 to a decimal

Anon25 [30]

Answer: 0.375

Step-by-step explanation:

5 0

3 years ago

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Area of the Region Between Curves-Please help.

Neporo4naja [7]

Answer: 32/3

Step-by-step explanation:

To find the area of the region between the curves, you first want to find the interval of both curves. You can do that by setting both curves equal to each other.

$-\frac{1}{2}x^2+2=\frac{1}{2} x^2-2$ [subtract both sides by 2]

$-\frac{1}{2}x^2= \frac{1}{2} x^2-4$ [subtract both sides by $\frac{1}{2} x^2$ ]

$-x^2=-4\\x=2,-2$

Now that we know the interval, we set this into an integral and subtract the top curve by the bottom curve. *Note: I can't type in -2 into the interval, therefore only a "-" will be displayed.

$\int\limits^2_- {|-\frac{1}{2}x^2+2-(\frac{1}{2}x^2-2) | } \, dx$ [distribute -1]

$\int\limits^2_-{|-\frac{1}{2}x^2+2-\frac{1}{2}x^2+2 |} \, dx$ [combine like terms]

$\int\limits^2_- {|-x^2+4|} \, dx$ [solve integral]

$\frac{32}{3}$

The area between the curves is 32/3. This was also the reason why we had the absolute values because area can never be negative.

6 0

4 years ago