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3 years ago

-K+ _ MgBr → _KBr + __Mg

Mathematics

1 answer:

Anit [1.1K]3 years ago

7 0

Answer:K + MgBr → KBr + Mg

This is an oxidation-reduction (redox) reaction:

MgI + e- → Mg0

(reduction)

K0 - 1 e- → KI

(oxidation)

MgBr is an oxidizing agent, K is a reducing agent.

Step-by-step explanation:

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If 2^5 x 5^2 = 2^n x 10^2, what is the value of n?

Marat540 [252]

Answer: x = 0

Step-by-step explanation:

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

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3 years ago

5x(6x+28)=-23 Help I don't understand

aksik [14]

Answer:

Solve the equation for x by finding a, b, and c of the quadratic then applying the quadratic formula.

Exact Form:

x = − 70 ± √4210/30

Decimal Form:

x = −0.03

Step-by-step explanation:

5x(6x+28)=−23

Step 1: Simplify both sides of the equation.

30x2+140x=−23

Step 2: Subtract -23 from both sides.

30x2+140x−(−23)=−23−(−23)

30x2+140x+23=0

Step 3: Use quadratic formula with a=30, b=140, c=23.

x=−b±√b2−4ac/2a

x=−(140)±√(140)2−4(30)(23)/2(30)

x=−140±√16840/60

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3 years ago

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Volume of right trapezoid cylindar whole bases are B 16m, b 8m, height is 4m and length is 32m

viktelen [127]

Answer:

$volume = 1536 m^3$

Step-by-step explanation:

given data;

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b =8 m

height H = 4 m

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Volume = A* L

The area of a trapezoid is

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3 years ago

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Answer:

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Step-by-step explanation:

The circumference of a circle with radius $r$ is given by $C=2\pi r$ . The length of an arc is makes up part of this circumference, and is directly proportion to the central angle of the arc. Since there are 360 degrees in a circle, the length of an arc with central angle $\theta^{\circ}$ is equal to $2\pi r\cdot \frac{\theta}{360}$ .

Formulas at a glance:

Circumference of a circle with radius $r$ : $C=2\pi r$
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Question 1:

The radius of the circle is 12 m. Therefore, the circumference is:

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In the circle shown, the radius is marked as 2 miles. Substituting $r=2$ into our circumference formula, we get:

$C=2(\pi)(2),\\C=\boxed{4\pi\text{ mi}}$

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$\ell_{arc}=4\pi \cdot \frac{135}{360},\\\ell_{arc}=1.5\pi=\boxed{\frac{3\pi}{2}\text{ mi}}$

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Step-by-step explanation:

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