All categories

3 years ago

How do I solve this problem (I did the graph)

Mathematics

1 answer:

Andrej [43]3 years ago

3 0

C = - 1/2t + 6
The equation usually looks like this

C = k*t + n

If t = 0 than

C = n (from the table we see that C is 6 if t = 0) than

N = 6 too

For t = 2 and c = 5 we have the equation

5 = 2k + 6
2k = -1
K = -1/2

You might be interested in

The first rule is subtract 6 starting from 48. The second rule is subtract 3 starting from 30. What is the second ordered pair u

Natasha2012 [34]

Answer:

(42,48)?

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Which choice is the equation of a line that passes through (9, -3) and is parallel to the line represented by this equation? y=5

miskamm [114]

Answer:

$y+3=\frac{5}{3}(x-9)$ or $y=\frac{5}{3}x-18$

Step-by-step explanation:

step 1

Find the slope of the line parallel to the given line

we know that

If two lines are parallel, then their slopes are the same

we have

$y=\frac{5}{3}x-4$

therefore

The slope m of the line parallel to the given line is

$m=\frac{5}{3}$

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{5}{3}$

$point\ (9,-3)$

substitute

$y+3=\frac{5}{3}(x-9)$ ----> equation of the line in point slope form

step 3

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$y+3=\frac{5}{3}(x-9)$

Isolate the variable y

$y+3=\frac{5}{3}x-15$

$y=\frac{5}{3}x-15-3$

$y=\frac{5}{3}x-18$

7 0

3 years ago

Of the smoothie sold yesterday at Isaac's smoothie shop 1/4 we're banana and another 2/3 were strawberry What fraction of smooth

lesya [120]

Answer:

11/12

Step-by-step explanation:

Fraction of smoothie that is Banana = 1/4

Fraction of smoothie that is strawberry = 2/3

The fraction of smoothies that were either banana or strawberry

= 1/4 + 2/3

Least Common Denominator= 12

= (3 × 1 + 2 × 4)/12

= 3 + 8/12

= 11/12

The fraction of smoothies were either banana or strawberry is 11/12

4 0

3 years ago

Which sequence of similar transformations could map △ABC onto △A'B'C'?

nadya68 [22]

Answer: A sequence of similar transformations of dilation and translation could map △ABC onto △A'B'C'.

Step-by-step explanation:

Similar transformations: If one figure can be mapped onto the other figure using a dilation and a congruent rigid transformation or a rigid transformation followed by dilation then the two figures are said to be similar.

In the attachment △ABC mapped onto △A'B'C' by a sequence of dilation from origin and scalar factor k followed by translation.