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sveticcg [70]
3 years ago
9

It Which is an equation of a line with a slope of 3 that passes through the origin?

Mathematics
2 answers:
Sonja [21]3 years ago
6 0
I have seen that question before i will try and find it and get back to you soon
gregori [183]3 years ago
5 0

Answer:

Payton, our equation looks like y = mx + b with m being the slope and b being the y-intercept. We are given a slope of 3, and if the graph passes through the origin, the y-intercept is 0

Step-by-step explanation:

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Lateral surface area
ivann1987 [24]
The answer is going to be <span>181.121506177483. hope that helped</span>
3 0
3 years ago
Iestion 10 (5 points)
marta [7]

Answer:

Jake is 9, and Jake's dad is 43.

Step-by-step explanation:

Ok, let's start off by putting a variable for Jack's age, A.

We know that Jack's father is 2 less than 5 times, so we can write his as 5x -2.

Now let's make an equation:

52 = x + 5x - 2

We need to isolate the x-values so we need to add 2 on both sides:

54= x + 5x

No we need to add the like terms:

54 = 6x

Finally, we just divide by 6 on each side.

Now we know X = 9 but we're not done yet :)

So because Jake's dad is almost 5 times as much, we first multiply Jack's age by 5. We get 45 but we need to subtract 2 which leaves us with 43.

3 0
3 years ago
Work out missing angle following polygons​
sineoko [7]

Answer:

x = 150°

Step-by-step explanation:

Interior angle of a hexagon = 120° and interior angle of a square = 90°

so remaining angle, 360-120-90 = 150°

7 0
3 years ago
7) y is inversely proportional to the square root of (x + 1).
Virty [35]
Mark Brainliest please

Answer y= 16/99

Explanation:
the initial statement is
y ∝ 1/x

To convert to an equation multiply by k the constant of variation
⇒ y = k × 1/x = k/x
To find k use the given condition y = 15 when x = 5

y = k/ x

⇒ k = y*x = 8 * 2 =16

equation is
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

y = 16/ x

when
x = 99
then

y = 16/99

Y = 16/99
3 0
3 years ago
What is 16% of 375?Thank You
Elan Coil [88]
The answer would be 60 hope this helps
6 0
3 years ago
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