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katen-ka-za [31]
3 years ago
9

Find the value of x in the equation below. 15.4=x-13.9

Mathematics
1 answer:
finlep [7]3 years ago
3 0

Answer:

29.3

Step-by-step explanation:

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help asap pls!!!!!!!!!! Which equation can be used to determine the distance between the origin and (–2, –4)?
Semenov [28]
We know that
the distance formula is
d=√[(y2-y1)²+(x2-x1)²]
 point (0,0) and point  <span>(–2, –4)
</span>d=√[(-4-0)²+(-2-0)²]---------> d=√[16+4]-------> d=√20
d=4.47 units

the answer is the option B
see the attached figure

8 0
3 years ago
Find the area of a parallelogram ABCD with AB =7 m. AD = 20 m and BAD = 62°.
nikklg [1K]

9514 1404 393

Answer:

  123.6 m²

Step-by-step explanation:

The area is given by the formula ...

  A = (AB)(AD)sin(∠A)

  A = (7 m)(20 m)sin(62°) ≈ 123.6 m²

4 0
3 years ago
What is the degree measure of one sixth of a right angle
Mariulka [41]
It is 15 degrees. A right angle is 90 degrees, so to find one sixth of that, you divide 90 by 6 (which is the same thing as multiplying 90 by one sixth). You get 15.
6 0
3 years ago
What is a postulate?
zloy xaker [14]

Answer:

A statement, also known as an axiom, which is taken to be true without proof. Postulates are the basic structure from which lemmas and theorems are derived.

Step-by-step explanation:

The Addition Postulate: If you have one apple and Sally has one apple, when you both add the same quantity to your existing number of apples, you'll still have the same number of apples. Using algebra, the postulate states:

If x = y, then x + 4 = y + 4

The Subtraction Postulate: If you have ten apples and Sally has ten apples, when you both subtract the same quantity of apples from your existing number of apples, you'll still have the same number of apples.

If x = y, then x - 3 = y - 3

Without being repetitive, these same principles apply to both multiplication and division.

The Multiplication Postulate: If x = y, then x * 3 = y * 3

The Division Postulate: If x = y, then x / 7 = y / 7

5 0
3 years ago
Read 2 more answers
Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\&#10;x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\&#10;x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\&#10;x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\&#10;(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;x\in(-\infty,-2\rangle\cup\langle3,\infty)


&#10;\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\&#10;x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\&#10;2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\&#10;\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\&#10;(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\&#10;(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\&#10;4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\
&#10;4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\&#10;(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\&#10;(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\&#10;(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\&#10;(x-1)^2(3x^2-28)=0\\&#10;x-1=0 \vee 3x^2-28=0\\&#10;x=1 \vee 3x^2=28\\&#10;x=1 \vee x^2=\dfrac{28}{3}\\&#10;x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\&#10;x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\&#10;\boxed{\boxed{x=1}}
3 0
3 years ago
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