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svet-max [94.6K]
2 years ago
12

What is the slope of the line that contains these points?

Mathematics
2 answers:
Anna35 [415]2 years ago
6 0

Answer:

the answer is -1 hope this helps ت︎シ︎☽︎❣︎

Anastasy [175]2 years ago
3 0

Answer:

-1 or 1/-1

Step-by-step explanation:

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Add.<br> 5/12 + 7/8<br> Write your answer as a fraction in simplest form.
Anastasy [175]
5/12=10/24 and 7/8= 21/24, add those and you will get 31/24 subtract 31 from24 and get 7 your new fraction is 1 7/24 that is 5/12 + 7/8 in simplest form
6 0
3 years ago
We study proofs in computing because it helps us understand data structures
alisha [4.7K]

Answer:

false

Step-by-step explanation:

we study proofs in computing because it helps us to understand the mathematical computation. With the help of proofs in computing it is very easy to understand all the mathematical computation but in given statement it is given that it helps us to understand data structure so it is a false statement.

8 0
3 years ago
The circumference of a circle is 119.32 kilometers. What is the circle's radius?
DochEvi [55]

Answer:

58.66 is the radius of the circle

4 0
2 years ago
Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=
Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = x^2 + y^2 +z^2 = 4^2

z^2 = 16 -x^2 -y^2

z = \sqrt{16-x^2-y^2}

The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}

Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

∴

dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to \sqrt{12}

dA = rdrdθ

∴

The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

= 8\pi ( - \sqrt{4} + \sqrt{16})

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0
3 years ago
Lucy is training for a long distance race. She runs 5/8 of a kilometre every day during her first week of training. How many kil
Vadim26 [7]

Answer:

here is the pic you wanted

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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