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3 years ago

Write the linear equation

Mathematics

1 answer:

valina [46]3 years ago

6 0

Answer:

y=-3/2x+4

slope=-3/2

y-int= 4

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Need help will give brainiest

e-lub [12.9K]

Answer:

3. x=50, y=80, 4. x=6.5, y=5

Step-by-step explanation:

I think, not entirely sure.

8 0

3 years ago

15/60 as a percentage

Ugo [173]

15/60=1/4=0.25
0.25*100=25
15/60 in % is 25%

3 0

3 years ago

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Whats the square root of 9/49. Please help il mark you as brainlist

scZoUnD [109]

Answer:

3/7

Step-by-step explanation:

the square root of 9 is 3, the square root of 49 is 7

5 0

3 years ago

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If the value of x is 3, then what is the value of 2-8x+4x+4

Veronika [31]

Hey there,
2 - 8x + 4x + 4
2 - 8(3) + 4(3) + 4
2 - 24 + 12 + 4
-6

Hope this helps :))

<em>~Top♥</em>

5 0

3 years ago

A quantity with an initial value of 6200 decays continuously at a rate of 5.5% per month. What is the value of the quantity afte

ELEN [110]

Answer:

410.32

Step-by-step explanation:

Given that the initial quantity, Q= 6200

Decay rate, r = 5.5% per month

So, the value of quantity after 1 month, $q_1 = Q- r \times Q$

$q_1 = Q(1-r)\cdots(i)$

The value of quantity after 2 months, $q_2 = q_1- r \times q_1$

$q_2 = q_1(1-r)$

From equation (i)

$q_2=Q(1-r)(1-r) \\\\q_2=Q(1-r)^2\cdots(ii)$

The value of quantity after 3 months, $q_3 = q_2- r \times q_2$

$q_3 = q_2(1-r)$

From equation (ii)

$q_3=Q(1-r)^2(1-r)$

$q_3=Q(1-r)^3$

Similarly, the value of quantity after n months,

$q_n= Q(1- r)^n$

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

$q_{48}=Q(1-r)^{48}$

Putting Q=6200 and r=5.5%=0.055, we have

$q_{48}=6200(1-0.055)^{48} \\\\q_{48}=410.32$

Hence, the value of quantity after 4 years is 410.32.

4 0

3 years ago

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