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3 years ago

15

List these in order from least to greatest. 0.25,0.1375,11/16, and 5/8

1 answer:

Effectus [21]3 years ago

4 0

LEAST TO GREATEST : 0.1375 — $\frac{5}{8}$ — $\frac{11}{16}$ — 0.25

EXPLANATION : 0.1375 will always come first here. It's literally the smallest one here. Moving on to the fractions, once you convert $\frac{5}{8}$ and $\frac{11}{16}$ and make the denominators similar, you would end up with $\frac{5}{8}$ = $\frac{10}{16}$ and $\frac{11}{16}$ (which stays the same.) When you compare the numerators, 10 is less than 11, so it comes first. And the greatest and most obvious is left, which is 0.25.

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John works 6.50 per hour. he worked 40.0 hours last week. what amput is on his check?

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Answer:

40 * 6.5 = 260 --> D. 260.00

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2 years ago

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What is the FDP of 0.3

exis [7]

Answer:

0.3 = 3/10 = 30%

Step-by-step explanation:

The reading of the number as "three tenths" tells you how to write the fraction:

0.3 = 3/10

You can also express the same value in hundredths. Of course /100 is just a long way to write %, so ...

0.3 = 0.30 = 30/100 = 30%

0.3 = 3/10 = 30%

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3 years ago

If sum of first 6 digits of AP is 36 and that of the first 16 terms is 255,then find the sum of first ten terms.

Cloud [144]

Answer:

100

Step-by-step explanation:

We have the sum of first n terms of an AP,

Sn = n/2 [2a+(n−1)d]

Given,

36= 6/2 [2a+(6−1)d]

12=2a+5d ---------(1)

256= 16/2 [2a+(16−1)d]

32=2a+15d ---------(2)

Subtracting, (1) from (2)

32−12=2a+15d−(2a+5d)

20=10d ⟹d=2

Substituting for d in (1),

12=2a+5(2)=2(a+5)

6=a+5 ⟹a=1

∴ The sum of first 10 terms of an AP,

S10 = 10/2 [2(1)+(10−1)2]

S10 =5[2+18]

S10 =100

This is the sum of the first 10 terms.

Hope it will help.

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2 years ago

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Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.

lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

$PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ$

Likewise:

$\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ$

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

$m\angle PAC=m\angle PAB+m\angle BAC$

Likewise:

$m\angle QAB=m\angle QAC+m\angle BAC$

Since ∠QAC ≅ ∠PAB:

$m\angle PAC=m\angle QAC+m\angle BAC$

And by substitution:

$m\angle PAC=m\angle QAB$

Thus:

$\angle PAC\cong \angle QAB$

Then by SAS Congruence:

$\Delta PAC\cong \Delta BAQ$

And by CPCTC:

$PC\cong BQ$

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2 years ago

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Verall luule<br> 8 x<br> 10²=

Mama L [17]

Answer: the answer would be 64

Step-by-step explanation:

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3 years ago