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3 years ago

6

You divide a number by 3,add 6 then subtract 7. The result is 4 . What is the number

2 answers:

Triss [41]3 years ago

4 0

The answer is 15, hope this helps

Kitty [74]3 years ago

3 0

The number and result is 15.

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Help me please so I can turn this in

Ber [7]

Answer:

10y-12

Step-by-step explanation:

3 0

3 years ago

Find the slope of the line represented by the table of values

schepotkina [342]

Answer:

The slope》<em><u>-5</u></em>

Step-by-step explanation:

hope it helps

have a great day!!

3 0

3 years ago

If a 5-pound bag of potatoes costs $5.25 , then how much would 7 pounds cost?

sashaice [31]

Answer:

7.35

Step-by-step explanation:

5 pounds -> 5.25

1 pound -> 1.05

7 pounds-> 7.35

4 0

3 years ago

The slope f′(x) at each point (x,y) on a curve y=f(x) is given, along with a point (a,b) on the curve. Use this information to f

Montano1993 [528]

$f'(x)=\dfrac{4x}{1+7x^2}$

Integrating gives

$f(x)=\displaystyle\int\frac{4x}{1+7x^2}\,\mathrm dx$

To compute the integral, substitute $u=1+7x^2$ , so that $\frac27\,\mathrm du=4x\,\mathrm dx$ . Then

$f(x)=\displaystyle\frac27\int\frac{\mathrm du}u=\frac27\ln|u|+C$

Since $u=1+7x^2>0$ for all $x$ , we can drop the absolute value, so we end up with

$f(x)=\dfrac27\ln(1+7x^2)+C$

Given that $f(0)=10$ , we have

$10=\dfrac27\ln1+C\implies C=10$

so that

$\boxed{f(x)=\dfrac27\ln(1+7x^2)+10}$

7 0

3 years ago

Someone please help me with this lol… have no idea what I’m doing

Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

$\sin \theta=-\sqrt{\dfrac{16}{25}}$

$\sin \theta=-\dfrac{4}{5}$

Therefore, the correct option is B.

6 0

2 years ago