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amid [387]
3 years ago
14

Please help on this one?

Mathematics
1 answer:
rodikova [14]3 years ago
4 0

Answer:

\large\boxed{A.\ \dfrac{4}{x-9}}

Step-by-step explanation:

\dfrac{4(x-2)}{(x-2)(x-9)}\\\\\text{canceled}\ (x-2)\\\\=\dfrac{4}{x-9}

You might be interested in
Given that ∠Ф = 180°, determine the numerical value of this function.
Rudiy27

Answer:

The answer to your question is 3. 0

Step-by-step explanation:

For an angle of 180°:

- hypotenuse = 1

- Adjacent side = 1

- Opposite side = 0

Then, sin Ф = \frac{Opposite side}{Hypotenuse}

         sin 180° = \frac{0}{1}

         sin 180° = 0

8 0
3 years ago
35 61 63 69 49 46 52 32 34 50<br> what is the mode
egoroff_w [7]

Answer:

no mode

Step-by-step explanation:

Mode is the most common number shown therefore....

since there are no numbers repeated, the answer is no mode

3 0
3 years ago
Can someone plz tell me the answer and how to do this math problem?
olasank [31]
Start by taking away 22.36 from each side so that you are left with 4.73c = 42.57
And then divide each side by 4.73 so you will get c = 9
5 0
3 years ago
Read 2 more answers
An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i
Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = \sqrt{(15-x)^2+1^2}

The total time, 't', is given as follows;

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

The minimum value of 't' is found by differentiating with an online tool, as follows;

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

6 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
Read 2 more answers
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