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DIA [1.3K]
2 years ago
11

Math graphs Find the graph

Mathematics
1 answer:
Nadya [2.5K]2 years ago
3 0

Step-by-step explanation:

put the value of y and get the value of x.then plot it in graph.(try to get non fractional values)

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Simplify: x + x + x + x + x + 2x *
Yuki888 [10]

Answer:

7x

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Select all ratios equivalent to 4:3.
valentina_108 [34]

Answer:

8:6 and 16:12

Step-by-step explanation:

5 0
3 years ago
Can someone please help me.
Katena32 [7]
Given 
- f(n) values for n=1,2,3,4
- possible candidates for the function

Solution:
Method: Evaluate some of the values, for each function.  A function with ANY value not matching the given f(n) values will be rejected.

N=1, f(n)=4
f(1)=4-3(1-1)=4
f(1)=4+3^(1+1)=4+3^2=4+9=13 ≠ 4   [rejected]
f(1)=4(3^(n-1))=4(3^0)=4
f(1)=3(4^(n-1))=3(4^0)=3*1=3  [rejected]

N=2, f(n)=12
f(1)=4-3(2-1)=4-3(1)=1 ≠ 12   [rejected]
[rejected]
f(1)=4(3^(2-1)=4*3^1=4*3=12
[rejected]

Will need to check one more to be sure
N=3, f(n)=3
[rejected]
[rejected]
f(3)=4(3^(n-1))=4(3^(3-1))=4(3^2)=4*9=36   [Good]
[rejected]

Solution: f(n)=4(3^(n-1))










6 0
3 years ago
Gronala costs $4.50 per pound. Assorted nuts cost $7.00 per pound. how many pounds of granola and nuts are needed to make 7 poun
yulyashka [42]

Let G be the granola we need (in pound)

Let N be the nut we need (in pound)

Trail mix cost $5.00 per pound---> cost for 7 pound = 5 x 7 = $ 35

4.5G+7N= 35

G+ N = 7

Therefore G = 5.6 N=1.4, we would need 5.6 pound granola & 1.4 pound nut

7 0
4 years ago
Read 2 more answers
A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s
Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

h(t)=-16t^2+32t+128

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

-16t^2+32t+128=0

The above is a quadratic equation. It can be solved by the formula as follows :

t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here, a = -16, b = 32 and c = 128

t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s

Neglecting negative value, the projectile will land after 4 seconds.

4 0
3 years ago
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