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scoray [572]
4 years ago
6

Is a ratio a rate always, sometimes, or never

Mathematics
1 answer:
dem82 [27]4 years ago
5 0
Ratio can be rate but rate can never be ratio
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10a - 25 + 56 need help asap​
Lady bird [3.3K]

Answer:

10a+31 is the simplified version but if you are trying to ask 10a-25=56 your answer would be 8.1

Step-by-step explanation:

5 0
3 years ago
The graph of quadratic function f(x) has a minimum at (-2,- 3 and passes through the point (2, 13). The functio
Nadusha1986 [10]

Answer:

Step-by-step explanation:

hello :  

the vertex -form of quadratic function f(x) is : f(x) = a(x+b)²+c

the vertex : (- b , c )  

means : (-2,-3)   so :  -b = -2   and  c = - 3      so  :  b =2  and  c = -3

f(x) = a(x+2)²-3

calculate : a  

the graph  passes through the point (2, 13) : f(2)=13

means : a(2+2)²-3 = 13

16a - = 13

16a = 16   so : a = 1

f(x) is represented by the equation f(x) = (x+2)²-3

The graph of  f(x) color red  and g(x) blue

is the y-intercept when g(x) greater g(x) than f(x) is when x=0  so : y = 6

8 0
3 years ago
What is 67% of 12? I NEED AN ANSWER QUICK.
professor190 [17]
The correct answer is C. you can get this by multilingual 67x0.12
4 0
3 years ago
Solving equations using Zero-Product property
FrozenT [24]

Answer:if y=0

Step-by-step explanation: view y as 0, so you have 2*0*(-8+0+7)=0

every number multiplied by 0 is 0 so 2*0 is 0 then in the brackets you will have -1. So the final line is 0*-1=0 like I said every number multiplied for 0 is 0 the equation is true for y=0

7 0
3 years ago
Read 2 more answers
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

4 0
3 years ago
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