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Sveta_85 [38]
2 years ago
15

What is an equation of the line that passes through the points (6, 4) and(6,−6)?

Mathematics
1 answer:
Tems11 [23]2 years ago
6 0

Answer:

Step-by-step explanation:

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In an isosceles triangle two sides are equal. The third side is two less than twice the length of the two equal sides. The perim
pogonyaev

Answer:

The equal sides' length = 10

Third side length = 18

Step-by-step explanation:

Let the two equal sides' length be "x"

Let the third side length be "y"

Third side (y) is 2 LESS THAN TWICE the equal side, so we can write:

y = 2x - 2

The perimeter is 38. The perimeters is the sum of all three sides. So we can write:

x + x + y = 38

Replacing y with 2nd equation, we have:

x + x + (2x - 2) = 38

Now, we can solve this for x first:

x + x + (2x - 2) = 38\\x+x+2x-2=38\\4x=38+2\\4x=40\\x=10

Now, y is:

y = 2x - 2

y = 2(10) - 2

y = 20 - 2

y = 18

The equal sides' length = 10

Third side length = 18

3 0
3 years ago
An element with mass 290 grams decays by 13.2% per minute. How much of the
Tatiana [17]

Answer:

FV=PV(1−d)^n

FV = 290(1-.132)^14

FV = 290(.868)^14

FV = 39.96 g

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Determine whether the measures 2, √8, and √12 can be the measures of the sides of a triangle. If so, classify the triangle as ac
gtnhenbr [62]

Answer:

C

Step-by-step explanation:

First check if the triangle is right.

If the square of the longest side is equal to the sum of the squares on the other 2 sides then the triangle is right.

longest side = \sqrt{12} and (\sqrt{12} )² = 12

2² + (\sqrt{8} )² = 4 + 8 = 12

Thus the triangle is right → C

7 0
2 years ago
Read 2 more answers
In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the
emmasim [6.3K]

Answer:

a=\frac{2S -2v_ot-2s_o}{t^2}

Step-by-step explanation:

We have the equation of the position of the object

S = \frac{1}{2}at ^2 + v_ot+s_o

We need to solve the equation for the variable a

S = \frac{1}{2}at ^2 + v_ot+s_o

Subtract s_0 and v_0t on both sides of the equality

S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o

S -v_ot-s_o = \frac{1}{2}at ^2

multiply by 2 on both sides of equality

2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2

2S -2v_ot-2s_o =at ^2

Divide between t ^ 2 on both sides of the equation

\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}

Finally

a=\frac{2S -2v_ot-2s_o}{t^2}

5 0
3 years ago
Who else going back to school on campus?
JulsSmile [24]

Heck naw Fawk that -3-!!!!!

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