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Stolb23 [73]
3 years ago
10

Mr. Bojorquez has 32 homework papers and 28 tests to return. Ms. Williams has 104 homework papers and 98 tests to return. Find e

ach teachers ratio of homework papers to the number of tests they have to return. Are the ratios equivalent?
Mathematics
2 answers:
Aleks04 [339]3 years ago
8 0

Answer:

No they are not equivalent

Step-by-step explanation:

The ratio for Ms.Bojorquez is 32:28 which simplified is 8:7

The ration for Ms.Williams is 104:98 which simplified is 52:49

These rations are not equivalent.

Inessa [10]3 years ago
4 0
No they are not equivalent
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7 is 5 greater than the 2
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4 years ago
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1. Mrs. Jones picks 3 flowers a day starting on Monday. How many does she have by Sunday?
Aleks04 [339]

Answer:

1. 18

2. 2412

Step-by-step explanation:

1. Monday 3

Tuesday 6

Wensday 9

Thursday 12

Friday 15

Sunday 18

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2.

603*4=2412

3 0
3 years ago
Sal bought x shares of a stock that sold for 31.50 per share. He paid a 1% commission on the sale. The total cost of his investm
Licemer1 [7]

Answer:

180

Step-by-step explanation:

add 1% to 31.50 to get 31.815

then take the total 5276.70 and divide by 31.815

5 0
3 years ago
HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270
BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

\large \boxed{sin \theta = tan \theta cos \theta} \\  \large \boxed{tan^{2}  \theta + 1 =  {sec}^{2} \theta}

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

\large{ {4}^{2}  + 1 =  {sec}^{2} \theta } \\  \large{16 + 1 =  {sec}^{2} \theta } \\  \large{ {sec}^{2}  \theta = 17}

As we know, sec²θ = 1/cos²θ.

\large \boxed{sec \theta =   \frac{1}{cos \theta} } \\  \large \boxed{ {sec}^{2}  \theta =  \frac{1}{ {cos}^{2}  \theta} }

And thus,

\large{  {cos}^{2}  \theta =  \frac{1}{17}}   \\ \large{cos \theta =  \frac{ \sqrt{1} }{ \sqrt{17} } } \\  \large{cos \theta =  \frac{1}{ \sqrt{17} }  \longrightarrow  \frac{ \sqrt{17} }{17} }

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

\large{cos \theta =   \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta =  -  \frac{ \sqrt{17} }{17}}

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

\large{sin \theta = 4 \times ( -  \frac{ \sqrt{17} }{17}) } \\  \large{sin \theta =  -  \frac{4 \sqrt{17} }{17} }

Answer

  • sinθ = -4sqrt(17)/17 or A choice.
4 0
3 years ago
I don't know how to solve it.
Nataly [62]

Step-by-step explanation:

Take the first derivative

\frac{d}{dx} ( {x}^{3}  - 3x)

3 {x}^{2}  - 3

Set the derivative equal to 0.

3 {x}^{2}  - 3 = 0

3 {x}^{2}  = 3

{x}^{2}  = 1

x = 1

or

x =  - 1

For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).

For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.

Since we are going to Positve to negative slope, we have a local max at x=-1

Plug in -1 for x into the original function

( - 1) {}^{3}  - 3(  - 1) = 2

So the local max is 2 and occurs at x=-1,

For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.

Since we are going to decreasing to increasing, we have minimum at x=1,

Plug in 1 for x into original function

{1}^{3}   - 3(1)

1 - 3 =  - 2

So the local min occurs at -2, at x=1

8 0
2 years ago
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