I need help with four questions NOW!!!! please.

How many different ways can you write a fraction that has a numerator of 2 as a sum of fractions? Explain.

zhenek [66]

Answer:

Step-by-step explanation:

1/5 + 1/5 = 2/5

1/7 + 1/7 = 2/7

1/3 + 1/3 = 2/3

There are an infinite number of these fractions. They must be 1 and 1 in the numerator, and the denominator must be relatively prime to 2. The examples I have picked are prime in the denominator, but the rule is not without many exceptions. For example

1/9 + 1/9 = 2/9

I don't think you can pick an even denominator because it will reduce when put with two. Oh wait 2/18 + 2/18 = 4/18 = 2/9 But these could be reduced before adding. Still, it might count. It depends on who is marking the question.

What about an odd and even denominator?

1/9 + 1/18 = 3/18 = 1/6 There must be something that works, but I can't come up with an example.

4 0

3 years ago

In ΔABC, ∠B measures 35° and the values of a and b are 19 and 11, respectively. Find the remaining measurements of the triangle,

Anna35 [415]

Answer:

a) ∠A = 82.2° , ∠C = 62.8° , c = 17.01

Step-by-step explanation:

Explanation:-

Step(i):-

Given data ∠B measures 35° and the values of a and b are 19 and 11

∠B = 35° and sides a = 19 and b = 11

By using sine rule

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{Sin C} = 2 R$

now we will use

$\frac{a}{sin A} = \frac{b}{sin B}$

$\frac{19}{sin A} = \frac{11}{sin 35}$

cross multiplication , we get

$\frac{19 X sin 35}{11} = sinA$

sin A = 0.990

A = sin⁻¹( 0.990) = 82.2°

∠A = 82.2°

Step(ii):-

we know that ∠A +∠B +∠C = 180°

∠C = 180° - ∠A -∠B

∠C = 180° -82.2°-35°

∠C = 62.8°

Step(iii):-

we will use formula

$\frac{b}{sin B} = \frac{c}{Sin C}$

$\frac{11}{sin 35} = \frac{c}{Sin 62.8}$

$\frac{11 X sin (62.8)}{sin 35} = C$

c = 17.01

Final answer:-

∠A = 82.2° , ∠C = 62.8° , c = 17.01

4 0

3 years ago

What is the PV of an ordinary annuity with 10 payments of $2,700 if the appropriate interest rate is 5.5%? Group of answer choic

Verizon [17]

Answer:

The PV of an ordinary annuity with 10 payments of $2,700 if the appropriate interest rate is 5.5% is $20,352.

Step-by-step explanation:

P = PMT [(1 - (1 / (1 + r) $^{n}$ )) / r]

= 2,700 [(1 - (1 / (1 + 0.055) $^{10}$ )) / 0.05]

= 2,700 [(1 - (1 / (1.708)) / 0.05]

= 2,700 [(1 - 0.58)) / 0.05]

= 2,700 [(0.41457) / 0.05]

= 2,700(7.53)

=$ 20,352

5 0

4 years ago

The graph shows that you are traveling at a constant rate. Three of the statements are true. Which is NOT?

Bess [88]

Answer:

Your speed is 135 miles per hour.

Step-by-step explanation:

Mark me brainliest if I helped:D

7 0

3 years ago

April worked 1 1/2 times as long on her math project as did Carl. Debbie worked 1 1/4 times as long as Sonia. Richard worked 1 3

vlada-n [284]

Answer:

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

Step-by-step explanation:

Some data's were missing so we have attached the complete information in the attachment.

Given:

Number of Hours Carl worked on Math project = $5\frac{1}{4}\ hrs$

$5\frac{1}{4}\ hrs$ can be Rewritten as $\frac{21}{4}\ hrs$

Number of Hours Carl worked on Math project = $\frac{21}{4}\ hrs$

Number of Hours Sonia worked on Math project = $6\frac{1}{2}\ hrs$

$6\frac{1}{2}\ hrs$ can be rewritten as $\frac{13}{2}\ hrs$

Number of Hours Sonia worked on Math project = $\frac{13}{2}\ hrs$

Number of Hours Tony worked on Math project = $5\frac{2}{3}\ hrs$

$5\frac{2}{3}\ hrs$ can be rewritten as $\frac{17}{3}\ hrs.$

Number of Hours Tony worked on Math project = $\frac{17}{3}\ hrs.$

Now Given:

April worked $1\frac{1}{2}$ times as long on her math project as did Carl.

$1\frac{1}{2}$ can be Rewritten as $\frac{3}{2}$

Number of Hours April worked on math project = $\frac{3}{2}$ $\times$ Number of Hours Carl worked on Math project

Number of Hours April worked on math project = $\frac{3}{2}\times \frac{21}{4} = \frac{63}{8}\ hrs \ \ Or \ \ 7\frac{7}{8} \ hrs$

Also Given:

Debbie worked $1\frac{1}{4}$ times as long as Sonia.

$1\frac{1}{4}$ can be Rewritten as $\frac{5}{4}$ .

Number of Hours Debbie worked on math project = $\frac{5}{4}$ $\times$ Number of Hours Sonia worked on Math project

Number of Hours Debbie worked on math project = $\frac{5}{4}\times \frac{13}{2}= \frac{65}{8}\ hrs \ \ Or \ \ 8\frac{1}{8}\ hrs$

Also Given:

Richard worked $1\frac{3}{8}$ times as long as tony.

$1\frac{3}{8}$ can be Rewritten as $\frac{11}{8}$

Number of Hours Richard worked on math project = $\frac{11}{8}$ $\times$ Number of Hours Tony worked on Math project

Number of Hours Debbie worked on math project = $\frac{11}{8}\times \frac{17}{3}= \frac{187}{24}\ hrs \ \ Or \ \ 7\frac{19}{24}\ hrs$

Hence We will match each student with number of hours she worked.

Student Hours worked

April. $7\frac{7}{8} \ hrs$

Debbie. $8\frac{1}{8}\ hrs$

Richard. $7\frac{19}{24}\ hrs$

5 0

3 years ago

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