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Vaselesa [24]
3 years ago
12

The coordinates of point T are (0,4). The midpoint of ST is (6,-7). Find the coordinate of point S

Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

Step-by-step explanation:

(x + 0)/2 = 6

x + 0 = 12

x = 12

(y + 4)/2 = -7

y + 4 = -14

y = -18

(12, -18)

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<img src="https://tex.z-dn.net/?f=%20%20%5Csqrt%7B2%20-%207x%7D%20%20%2B2x%20%3D%200" id="TexFormula1" title=" \sqrt{2 - 7x} +
FrozenT [24]

Answer:

√(2 - 7x) + 2x = 0

√(2 - 7x) = - 2x

[√(2 - 7x)]² = (- 2x)²

2 - 7x = 4x²

4x² + 7x - 2 = 0

4x² + 8x - x - 2 = 0

4x(x + 2) - (x + 2) = 0

(4x - 1)(x + 2) = 0

4x - 1 = 0 => 4x = 1 => x₁ = 1/4

x + 2 = 0 => x₂ = - 2

7 0
3 years ago
Find the slope from the graph<br><br> 5<br> -5<br> undefined<br> 0
yanalaym [24]

Answer:

0

Step-by-step explanation:

5-5=0

I hope this is correct and have a great day

8 0
3 years ago
Forth grade math: What happens to the area of the triangle when one of the dimensions is doubled and halved?
Alja [10]

\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ b=\stackrel{doubled}{2b} \end{cases}\implies A=\cfrac{1}{2}(2b)(h)\implies \stackrel{\textit{twice as the original area}}{A=bh} \\\\[-0.35em] ~\dotfill

\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ h=\stackrel{halved}{\frac{h}{2}} \end{cases}\implies A=\cfrac{1}{2}b\left( \cfrac{h}{2} \right)\implies \stackrel{\textit{half of the original area}}{A=\cfrac{1}{2}bh\left( \cfrac{1}{2} \right)}

7 0
4 years ago
Can one of you guys pls help
Gwar [14]

Answer:

Puts the dots where the A,B,C,D dots are and move them 5 units to the right.

Example:

Original place

A=(2,5)

Moved 4 units to the right:

A'=(6,5)

Here are some tips:

Up= Y value goes up

Down= Y value goes down

Right= X value goes up

Left X value goes down

6 0
3 years ago
Quadratic formula 3x^2 - 2x + 5 = 0
Tanzania [10]

Step-by-step explanation:

The given quadratic equation is :

3x^2 - 2x + 5 = 0

We need to solve the given equation. The formula used to solve a quadratic equation is :

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

We have, a = 3, b = -2 and c = 5

So,

x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(3)(5)}}{2(3)}\\\\x=0.333,4.667

Hence, this is the required solution.

4 0
3 years ago
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