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Elena-2011 [213]
4 years ago
5

HELP ! I have math soon

Mathematics
1 answer:
frosja888 [35]4 years ago
6 0
X=12 or 12+x or x+12 let me know if that gave u a idea and helped thanks

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Identify the solution to the system of equations <br> {x=-2y-6 -3x-7y=18
Gemiola [76]

Answer:

(- 6, 0 )

Step-by-step explanation:

x = -2y - 6 → (1)

- 3x - 7y = 18 → (2)

substitute x = - 2y - 6 into (2)

- 3(- 2y - 6) - 7y = 18 ← distribute parenthesis and simplify left side

6y + 18 - 7y = 18

- y + 18 = 18 ( subtract 18 from both sides )

- y = 0 , then

y = 0

substitute y = 0 into (1)

x = - 2(0) - 6 = 0 - 6 = - 6

solution is (- 6, 0 )

7 0
3 years ago
Read 2 more answers
Which of the diagram below represents the statement. if it is an insect then it has wings
ioda
The diagram that represent the statement is a Venn diagram in which:

An inner circle labeled "insects" is completely included inside an outer circle which is labeled "has wings".

In such diagram all the insects are inside the bigger set of things that have wings. That means that all the insects have wings but not all the things that have wings are insects.

That is exactly what the proposition if it is an insect then it has wings.

As per the proposition, having wings is a necessary condition to be insect, but it is not sufficient.
8 0
3 years ago
Use the quadratic formula to solve x^2 -3x - 5 = 0. Round to two decimal places.
Evgen [1.6K]

Answer:

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{3+\sqrt{29}}{2},\:x=\frac{3-\sqrt{29}}{2}

Step-by-step explanation:

Given the equation

x^2\:-3x\:-\:5\:=\:0

solving with the quadratic formula

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=-3,\:c=-5

x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}

x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{29}}{2\cdot \:1}

separating the solutions

x_1=\frac{-\left(-3\right)+\sqrt{29}}{2\cdot \:1},\:x_2=\frac{-\left(-3\right)-\sqrt{29}}{2\cdot \:1}

solving

x=\frac{-\left(-3\right)+\sqrt{29}}{2\cdot \:\:1}

  =\frac{3+\sqrt{29}}{2\cdot \:1}

  =\frac{3+\sqrt{29}}{2}

also solving

\:x=\frac{-\left(-3\right)-\sqrt{29}}{2\cdot \:\:1}

  =\frac{3-\sqrt{29}}{2\cdot \:1}

  =\frac{3-\sqrt{29}}{2}

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{3+\sqrt{29}}{2},\:x=\frac{3-\sqrt{29}}{2}

7 0
3 years ago
The model below represents an equation. Write the equation that is modeled then find what value of x makes the equation true.
Dimas [21]
6x-5=4x+5
+5

6x=4x + 0
-4x
2x=0
x =0

probably to late but whatever

4 0
3 years ago
The triangles are isosceles and ΔABC : ΔJKL. What is the length of J-L?
Tatiana [17]
You know ABC is isoceles, so AB = BC and BC = 2.5 cm.
That means that the scale factor is KL/BC = 2
So JL = 2AC = 8 cm
5 0
4 years ago
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