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ch4aika [34]
3 years ago
13

-the equations x-y=-14 and -x-y=14 what is the number of solutions? A.) no solutions B.) infinitely many solutions C.) one solut

ion
Mathematics
1 answer:
Yuki888 [10]3 years ago
6 0

Answer:

The answer is C one solution

Step-by-step explanation:

x-y=-14

-y=-x-14  bring x to the other side

y=x+14   divide everything by -1

-x-y=14

-y=x+14  bring x to the other side

y=-x-14  divide everything by -1



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How many points need to be removed from this graph
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3 is your answer mate
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3 years ago
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Let ? = {a, b, c} be a sample space. let m(a) = 1/2, m(b) = 1/3, and m(c) = 1/6. find the probabilities for all eight subsets of
Leviafan [203]
<span>
A= {a,b,c} . Since this set has 3 elements, the number
of its total subset is 2³ = 8 (including the Ф element):

Here below all the subsets of {a,b,c}, with their related probabilities, knowing that P(a) = 1/2 ; P(b) = 1/3 and P(c) = 1/6

{a}        </span>→→→→1/2
<span>{b}        </span>→→→→1/3
<span>{c}        </span>→→→→1/6
<span>{a,b}     </span>→→→→1/2 + 1/3 =  5/6
<span>{a,c}     </span>→→→→1/2 + 1/6  =  2/3
<span>{b,c}     </span>→→→→1/3 + 1/6  =  1/2
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5 0
3 years ago
Un cono ha l'area laterale di 255 pigreco cm^2, l'apotema di 17 cm e pesa 900 pigreco g. Calcola il peso specifico del materiale
valkas [14]

Answer:

The specific weight is 1.5\frac{g}{cm^{3}}

Step-by-step explanation:

The question in English

A cone has a lateral area of 255 pi cm^2, an apothem of 17 cm and weighs 900 pi g. It calculates the specific weight of the material of which it is composed

step 1

Find the radius of the cone

we know that

The lateral area of a cone is equal to

LA=\pi rl

we have

LA=255\pi\ cm^{2}

l=17\ cm

substitute the values

255\pi=\pi r(17)

Simplify

255=r(17)

r=255/(17)=15\ cm

step 2

Find the height of the cone

Applying the Pythagoras Theorem

l^{2} =r^{2} +h^{2}

substitute the values and solve for h

17^{2} =15^{2} +h^{2}

h^{2}=17^{2}-15^{2}

h^{2}=64

h=8\ cm

step 3

Find the volume of the cone

The volume of the cone is equal to

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

substitute the values

V=\frac{1}{3}\pi (15)^{2}(8)

V=600\pi\ cm^{3}

step 4

Find the specific weight

Divide the mass by the volume

\frac{900\pi }{600\pi}=1.5\frac{g}{cm^{3}}

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