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2 years ago

Which equations represent a line that passes through the points given in the table? Check all that apply.

Mathematics

2 answers:

solong [7]2 years ago

5 0

Answer:

I believe the answer is the 2nd box{y-2=1/6 (x+10)}, 3rd box{y-1=1/6(x+4)}, and the 5th box{y=1 x/6 +1/3}.

Step-by-step explanation:

anastassius [24]2 years ago

3 0

Answer: 2, 3, and 5.

Step-by-step explanation: Graph in pictures

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Which expression gives the length of the transverse axis of the hyperbola

klio [65]

Answer:

Option (B)

Step-by-step explanation:

From the picture attached,

F1 and F2 are the focii of the hyperbola.

Point P(x, y) is x units distant from F1 and y units distant from the other focus F2.

By the definition of a hyperbola,

"Difference between the distances of a point from the focii is always constant and equals to the measure of transverse axis."

Difference in the distances of point P from focii F1 and F2 = (x - y) units

This distance is equal to the length of the transverse axis = (x - y) units

Therefore, Option (B) will be the answer.

7 0

2 years ago

Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

Bumek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

is dimensionally consistent

is not dimensionally consistent

is dimensionally consistent

is not dimensionally consistent

is dimensionally consistent

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

8 0

2 years ago

Maya says that value of the 6 in 0.006 is 1/10 of the value of the 6 in 0.6 . Is she correct? Explain

Liono4ka [1.6K]

Answer:

Maya is incorrect

Step-by-step explanation:

1/10 of 0.6=

0.1*0.6=0.06

0.06≠0.006

Maya is incorrect

4 0

2 years ago

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OLga [1]

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2 years ago

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Keenan gives Tisha half of his strawberries. Tisha keeps 4 of the strawberries she hit from Keenan and gives the rest to suvi. W

Darya [45]

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8 0

2 years ago