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IRISSAK [1]
2 years ago
5

Plz help. I dont know how to solve this plz.

Mathematics
2 answers:
Shkiper50 [21]2 years ago
7 0

Answer:

its 0,6

Step-by-step explanation:

you just move however much it asks

Oliga [24]2 years ago
4 0

Answer:

(0,6)

Step-by-step explanation:

your basically imagining that your moving that point following the directions in the word problem. go left 3 times, go up 2 times.

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7 1/7 divided by 5 1/5
vladimir1956 [14]

Well first turn 1/7 and 1/5 into decimals which would be .142857 and .2 Add those to 7 and 5 to get 7.143 (I rounded this to the nearest thousandth) and 5.2 Divide those and you get 1.374

5 0
3 years ago
. Use the number line to illustrate the difference: -3-(-7)​
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3 years ago
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Nuetrik [128]
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4 0
3 years ago
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For what value does x does 3^2x= g^3x-4<br> A)1<br> B)2<br> C)3<br> D)4
netineya [11]
The question has an error because the letter g does not make sense in the context.

I will assume that the g is really the number 9.

In that case, the equation to solve would be:

3^{2x} =  9^{3x-4}

You can solve for x following these steps:

1) make      9=3^2

=>

3^{2x} = 3^{2(3x-4)}

2) Given that the basis are equal the exponents have to be equal =>

2x = 2(3x - 4)

3) Solve:

2x = 6x - 8

6x - 2x = 8

4x = 8

x = 8/4

x = 2 which is the option B) which leads me to think that a 9 instead of g in the equation should be right.

Under that assumption, the answer is the option B) x = 2.

3 0
3 years ago
What is the equation of a parabola with (−2, 4) as its focus and y = 6 as its directrix? Enter the equation in the box.
SOVA2 [1]
Notice the picture below

the directrix is above the focus point, meaning the parabola is vertical and is opening downwards

now, "p" is the distance from the vertex to the focus point or the directrix, so that means, the vertex is between those two fellows, over the axis of symmetry, x = -2, since "p" is 1 unit, that puts the vertex at -2,5

since the parabola is opening downwards, that means the "p" value is negative, so is -1

\bf \textit{parabola vertex form with focus point distance}\\\\&#10;\begin{array}{llll}&#10;(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\&#10;\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ({{ h}},{{ k}})\\\\&#10;{{ p}}=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\&#10;-----------------------------\\\\

\bf \begin{cases}&#10;p=-1\\&#10;h=-2\\&#10;k=5&#10;\end{cases}\implies (x-(-2))^2=4(-1)(y-5)&#10;\\\\\\&#10;(x+2)^2=-4(y-5)\implies &#10;-\cfrac{1}{4}(x+2)^2=y-5&#10;\\\\\\&#10;\boxed{-\cfrac{1}{4}(x+2)^2+5=y}

4 0
3 years ago
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