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3 years ago

Y=5x+2 what point at the y axis

1 answer:

5 0

I think to find the y - axis you should sub x=0
Therefore, y=5(0)+2
=2
Hence the answer is (0,2)

I'm sorry if this is wrong!

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Which choices are equivalent to the expression below? Check all that apply.

gavmur [86]

There’s no answer choices..

5 0

2 years ago

What is the slope of this graph? −1/3 3 1/3 −3

FromTheMoon [43]

I'm assuming there is probably a graph that goes with this problem so here is all the info I can give you with the four choices: If it is -1/3, the line will go DOWN 1, over 3 (just count on the graph) If it is 3, the line will go UP 3, over 1 If it is 1/3, the line will go UP 1, over 3 And if it it -3, it will go DOWN 3, over 1 (If the slope is 3 or -3, it would be quite steep compared to a slope of 1/3 or -1/3) Hope that helps!

4 0

3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

The map distance from Dodgetown to Kidville is 2.75 cm. Find the actual distance from Dodgetown to Kidville.

Fantom [35]

The answer is:

55 km (a)

7 0

3 years ago

Read 2 more answers

I NEED HELP ASAP

Allushta [10]

Euler's formula is given by:
C + V = A + 2
Where,
C: number of faces
V: number of vertices
A: number of edges.
Clearing A we have:
A = C + V-2
Substituting values:
A = 21 + 14-2
A = 33
Answer:
the missing number is:
C.33

7 0

3 years ago