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mixer [17]
4 years ago
7

Simplify 2 times the 4th root of 80

Mathematics
2 answers:
xz_007 [3.2K]4 years ago
8 0
The answer is <span>4 times the 4th root of 5. You would take out a 16, and the fourth root of that is 2 so 2 x 2 = 4 which would be outside the fourth root of 5.

Hope this helps! :)</span>
marishachu [46]4 years ago
3 0

Answer:

The correct option is A) 4 times the 4th root of 5 or 4\sqrt[4]{5}.

Step-by-step explanation:

Consider the provided information.

Simplify 2 times the 4th root of 80

This can be written as:

2\sqrt[4]{80}

2\sqrt[4]{80}

The above expression can be written as:

2\sqrt[4]{2 \times 2 \times 2 \times 2 \times 5}

2 \times 2\sqrt[4]{5}

4\sqrt[4]{5}

Thus, the correct option is A) 4 times the 4th root of 5 or 4\sqrt[4]{5}.

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the area of a circle is 96 square inches. What is the area of a circlewhose diameter is 1.5 times the diameter of the circle?
mars1129 [50]
The area of a circle A equals either:
πr² or πd²/4

96 = πd²/4 => d = 11 inches

The diameter of the second circle equals :
11*1.5 = 16.5 inches

The area equals: π(16.5)²/4 = 214 square inches

Good luck
5 0
4 years ago
I really don't understand this.
prohojiy [21]

Answer:

parang wla po yung answer

Step-by-step explanation:

8 0
3 years ago
What is the sign of -201 + 189−201+189minus, 201, plus, 189?
svlad2 [7]
-201+189=-12-201=-213+189=-24 . S the sign would be negative(-).
8 0
3 years ago
Carlos finished 1/3 of his art project on Monday. Tyler finished 1/2 of his art project on Monday. Who finished more of hood art
vladimir2022 [97]
Tyler finished more of the art project.

Carlos finished 1/3
In terms of percentage 1/3 *100% = 33.3%

Tyler finished 1/2
In terms of percentage 1/2 *100% = 50%

50% > 33.3%
50% is greater than 33.3%
4 0
3 years ago
Read 2 more answers
Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and
Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

\angle EAC = \angle ACD = 90^{\circ}

To prove :

EB\times BD=AB\times BC

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

\triangle AEC \sim \triangle EBA,

Similarly,

\triangle AEC \sim \triangle ABC

\implies\triangle EBA\sim \triangle ABC-----(1)

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

\triangle ADC \sim \triangle CBD,

Similarly,

\triangle ADC \sim \triangle ABC

\implies \triangle CBD\sim \triangle ABC-----(2)

From equations (1) and (2),

\triangle EBA\sim \triangle CBD

The corresponding sides of similar triangles are in same proportion,

\frac{EB}{BC}=\frac{AB}{BD}

\implies EB\times BD=AB\times BC

Hence, proved....

5 0
3 years ago
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