Answer:
1,5,25,125,
all can be added or multiplied to equal 125
The equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12) is y = -5x + 1
<h3>Equation of a line</h3>
A line is the distance between two points
Given the equation of a line expressed as 10x + 2y = -2. Determine the slope
2y = -10x -2
y = -5x - 1
Slope of the line is -5
The equation of a line in point-slope form is y - y1 = m(x-x1)
Substitute the point and the slope of the parallel line
y - 12 = -5(x - 0)
y - 12 = -5x
y = -5x + 12
Hence the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12) is y = -5x + 12
Learn more on equation of a line here: brainly.com/question/13763238
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<u>Solution-</u>
Given that,
In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.
Then considering ΔPQT and ΔSTF,
1- ∠FTS ≅ ∠PTQ ( ∵ These two are vertical angles)
2- ∠TFS ≅ ∠TPQ ( ∵ These two are alternate interior angles)
3- ∠TSF ≅ ∠TQP ( ∵ These two are also alternate interior angles)
<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>
∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.
As QS = TQ + TS = 10 (given)
If TS is x, then TQ will be 10-x. Then putting these values in the equation
∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm
Mass = 100 g
Volume = 50 cm³
Density :
D = m / V
D = 100 / 50
D = 2 g/cm³
answer A
hope this helps!.
<span>You can probably just work it out.
You need non-negative integer solutions to p+5n+10d+25q = 82.
If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80.
So this is the same as n + 2d + 5q ≤ 16
So now you simply have to "crank out" the cases.
Case q=0 [ n + 2d ≤ 16 ]
Case (q=0,d=0) → n = 0 through 16 [17 possibilities]
Case (q=0,d=1) → n = 0 through 14 [15 possibilities]
...
Case (q=0,d=7) → n = 0 through 2 [3 possibilities]
Case (q=0,d=8) → n = 0 [1 possibility]
Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81
Case q=1 [ n + 2d ≤ 11 ]
Case (q=1,d=0) → n = 0 through 11 [12]
Case (q=1,d=1) → n = 0 through 9 [10]
...
Case (q=1,d=5) → n = 0 through 1 [2]
Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42
Case q=2 [ n + 2 ≤ 6 ]
Case (q=2,d=0) → n = 0 through 6 [7]
Case (q=2,d=1) → n = 0 through 4 [5]
Case (q=2,d=2) → n = 0 through 2 [3]
Case (q=2,d=3) → n = 0 [1]
Total from case q=2: 1 + 3 + 5 + 7 = 16
Case q=3 [ n + 2d ≤ 1 ]
Here d must be 0, so there is only the case:
Case (q=3,d=0) → n = 0 through 1 [2]
So the case q=3 only has 2.
Grand total: 2 + 16 + 42 + 81 = 141 </span>
