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3 years ago

What's one way to check the answer to 5×6=30? a.30÷6=5 b.30×6=5 c.30×5=6 d.5÷6=30

Mathematics

2 answers:

Norma-Jean [14]3 years ago

7 0

Answer:

Option a -30÷6=5

Step-by-step explanation:

Given : Expression 5×6=30

Now , we check one by one which one gives the same result or a one way to answer the expression

a. 30÷6=5 is the one way because

5×6=30

⇒5=30÷6 which is true

b. 30×6=5 is not the correct expression because 30×6=180 not 5

c. 30×5=6 is not the correct expression because 30×5=150 not 6

d. 5÷6=30 is not the correct expression because 5÷6=0.833 not 30

Therefore, one way to check the answer is option a

VikaD [51]3 years ago

3 0

Solve is in the picture

Good luck :)

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$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
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(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

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

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

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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3 years ago

AB is a diameter of a circle, center O. C is a point on the circumference of the circle, such that

Solnce55 [7]

Given:

AB is the diameter of a circle.

m∠CAB = 26°

To find:

The measure of m∠CBA.

Solution:

Angle formed in the diameter of a circle is always 90°.

⇒ m∠ACB = 90°

In triangle ACB,

Sum of the angles in the triangle = 180°

m∠CAB + m∠ACB + m∠CBA = 180°

26° + 90° + m∠CBA = 180°

116° + m∠CBA = 180°

Subtract 116° from both sides.

116° + m∠CBA - 116° = 180° - 116°

m∠CBA = 64°

The measure of m∠CBA is 64°.

7 0

3 years ago