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jenyasd209 [6]
3 years ago
14

What is the quotient 5^-6 over 5^3

Mathematics
1 answer:
wel3 years ago
7 0

Answer:

1/ 5^9

Step-by-step explanation:

5^-6/5^3

5^(-6-3)

5^-9

Final Answer - 1/5^9

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maria brought a blouse on sale for 20% off. the sale price was $21.76. what was the original price? round to the nearest hundred
Degger [83]

Answer: $27.20

Step-by-step explanation:

She bought a blouse for 20% off so it means that she paid 80% of the original price.We can then set up the equation  

80% of x = 21.76   where x is the original price so solve for x

0.8x = 21.76  

x = 27.20

Which  means the original price is $27.20

4 0
3 years ago
15. Robin ordered pizza and opted for delivery.
Ivenika [448]

Answer: No she'll need atlease 9 dollars

Step-by-step explanation:

48.95 x 18% is 8.811 which is greater than

4 0
2 years ago
a total of 695 tickets were sold for the school play. They were either adult or student tickets. There was 55 fewer students tic
FinnZ [79.3K]

Answer:

375 + 320

Step-by-step explanation:

375 adult

320 students

equals 395

4 0
2 years ago
What is the answer to 6+9y-18=-3
lara [203]
6 + 9y - 18 = -3   Combine like terms (6 and -18)
      9y - 12 = -3   Add 12 to both sides
             9y = 9    Divide both sides by 9
               y = 1
7 0
3 years ago
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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