All categories

3 years ago

Please Help!!! 25 points!

Mathematics

2 answers:

iren [92.7K]3 years ago

6 0

Answer:.

Step-by-step explanation:

julia-pushkina [17]3 years ago

3 0

Answer: B/E

Step-by-step explanation:

Because the reciprocal are congruent to both sides

You might be interested in

The angles of a triangle are 128, 5x+10 and 9x. What is the value of x?

Advocard [28]

Angles in a triangle total 180°

180=128+5x+10+9x
180-128-10=5x+9x
42=14x
x=3

5 0

3 years ago

Read 2 more answers

Can someone please check this (#4) for me?

IrinaVladis [17]

Answer:

What you are doing wrong is that you are using the formula to find the area of the entire circle rather than using the formula to find a sector of a circle. The two cases uses 2 different kind sentences of formulas.

3 0

3 years ago

PLEASSSSEEEEE I AM SOOO CONFUSED BECAUSE IT SHOWS THE CORRECT ANSWER BUT I DONT GET ITTTTTT

IceJOKER [234]

Answer:

D. {(2,3) (2,5) (4,7)}

Step-by-step explanation:

2 has an output od 3 and 5, and 4 has an output of 7

7 0

2 years ago

Read 2 more answers

The amount of radioactive element remaining, r, in a 100mg sample after d days is represented using the equation r=100(1/2) d/5.

Ludmilka [50]

Answer:

12.94%

Step-by-step explanation:

r = 100(1/2)^(d/5) = 100((1/2)^(1/5))^d ≈ 100(.87055)^d

The daily decrease is 1 - 0.87055 = 0.12944 ≈ 12.94%

4 0

3 years ago

One container is filled with mixture that is 25% acid. A second container is filled with a mixture that is 55% acid. The second

algol13

Answer:

Therefore the third container contains $\frac{735}{17}\%$ = 43.23 % acid.

Step-by-step explanation:

Given that, One container is filled with the mixture that 25% acid. 55% acid is contain by second container.

Let the volume of first container be x cubic unit.

Since the volume of second container is 55% larger than the first.

Then the volume of the second container is

$= (V\times \frac{100+55}{100})$ cubic unit.

$=\frac{155}{100}V$ cubic unit.

The amount of acid in first container is

$=V\times \frac{25}{100}$ cubic unit.

$=\frac{25}{100}V$ cubic unit.

The amount of acid in second container is

$=(\frac{155}{100}V \times \frac{55}{100})$ cubic unit.

$=\frac{8525}{10000}V$ cubic unit.

Total amount of acid $=(\frac{25}{100}V+\frac{8525}{10000}V)$ cubic unit.

$=(\frac{2500+8525}{10000}V)$ cubic unit.

$=(\frac{11025}{10000}V)$ cubic unit.

Total volume of mixture $=(V+\frac{155}{100}V)$ cubic unit.

$=\frac{100+155}{100}V$ cubic unit.

$=\frac{255}{100}V$ cubic unit.

The amount of acid in the mixture is

$=\frac{\textrm{The volume of acid}}{\textrm{The amount of mixture}}\times 100 \%$

$=\frac {\frac{11025}{10000}V}{\frac{255}{100}V}\times 100\%$

$=\frac{735}{17}\%$

Therefore the third container contains $\frac{735}{17}\%$ acid.

5 0

3 years ago