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2 years ago

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Which equation in standard form has a graph that passes through the point (-2, 6) and has a slope of !.A. 3x — 2y = 12 B. 3x — 2

y = 6 C. 3x — 2y = -18 D. 3x — 2y = 18

2 answers:

pashok25 [27]2 years ago

7 0

Answer:

3x - 2y = -18

Step-by-step explanation:

3x - 2y = -18 and (-2, 6)

(3)(-2) - (2)(6) = -6 - 12 = -18

Keith_Richards [23]2 years ago

4 0

Answer:

Your answer should be C

Step-by-step explanation:

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A set of equations is given below:

ipn [44]

The correct answer is 'Equation F can be written as 2d + 1 = 3d + 7'. In order to find the answer to a system of equations, the two equations must be set equal to each other. For instance, if we had the equations x = y + 1 and x = 3y - 1, we would set the two equations equal to one another to find the answer.

x = y + 1
x = 3y - 1
y + 1 = 3y - 1
1 = 2y - 1
2 = 2y
1 = y

x = y + 1
x = 1 + 1
<span>x = 2

We can use this to solve the set of equations above.

</span><span>2d + 1 = 3d + 7
</span>1 = d + 7
-6 = d

c = 2d + 1
c = 2(-6) + 1
c = -12 + 1
c = -11

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3 years ago

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Find the value of the variable if P is between J & K. A) JP = 8z - 17 B) PK = 5z + 37 C) JK = 17z - 4

Daniel [21]

<span>We cannot deduce about the exact location of P between J and K. But we can conclude: segment JP + segment PK = line JK.
</span><span>JP + PK = JK.
</span><span>Substitute first each.
(8z - 17) + (5z + 37) = 17z - 4
Combine like terms.
13z + 20 = 17z - 4
Isolate the variable z.
4z = 24
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The value of the variable z is then 6 units.</span>

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If the ratio of dogs to cats at an animal shelter is 2:5, then the number of cats is _____ times the number of dogs.

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The answer is 2.5. Just make 5 and 2 0.4 as a ratio

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3 years ago

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Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

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german

3 = 48 divided by 16

Mark brainliest please

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