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4 years ago

Can you please help me you on this

Mathematics

1 answer:

Lana71 [14]4 years ago

5 0

Answer:on what?

Step-by-step explanation:I can’t see it

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Colleen McHugh wishes to deposit checks for $13.75, $92.08, and $8.21 into her account. She would like to receive $15.00 in cash

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her total deposit is 98.94

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3 years ago

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Makani receives a weekly allowance for doing chores around the house. Makani saves his money for 13 weeks. After 13 weeks, he sa

steposvetlana [31]

Answer:

divide amount by number of weeks.

156/13 = 12

$12

4 0

3 years ago

Find the slope of each line and then determine if the lines are parallel, perpendicular or neither. If a value is not an integer

Aleonysh [2.5K]

As given by the question

There are given that the point of two-line

$\begin{gathered} (1,\text{ 7) and (5, 5)} \\ (-1,\text{ -3) and (1, 1)} \end{gathered}$

Now,

From the condition of a parallel and perpendicular line

If the slopes are equal then the lines are parallel

If the slopes are negative reciprocal then the lines are perpendicular

If the slopes are neither of the above are true then lines are neither

Then,

First, find the slope of both of line

So,

For first-line, from the formula of slope

$\begin{gathered} m_1=\frac{y_2-y_1}{x_2-x_1} \\ m_1=\frac{5_{}-7_{}}{5_{}-1_{}} \\ m_1=-\frac{2}{4} \\ m_1=-\frac{1}{2} \end{gathered}$

Now,

For second-line,

$\begin{gathered} m_1=\frac{y_2-y_1}{x_2-x_1} \\ m_1=\frac{1_{}-(-3)_{}}{1-(-1)_{}} \\ m_1=\frac{4}{2} \\ m_1=2 \end{gathered}$

The given result of the slope is negative reciprocal because

$\begin{gathered} -\frac{1}{2}=-(-\frac{2}{1}) \\ -\frac{1}{2}=2 \end{gathered}$

Hence, the slope of line1 is -1/2, and slope of line2 is 2 and the lines are perpendicular.

5 0

1 year ago

Can you use distributive property for 567÷9

kramer

No because distributive property only for multiplication

6 0

4 years ago

Read 2 more answers

Prove that: (secA-cosec A) (1+cot A +tan A) =( sec^2A/cosecA)-(Cosec^2A/secA)<br>

Ksju [112]

Step-by-step explanation:

$(\sec A - \csc A)(1 + \cot A + \tan A)$

$=(\sec A - \csc A)\left(1 + \dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A} \right)$

$=(\sec A - \csc A)\left(1 + \dfrac{\cos^2 A + \sin^2 A}{\sin A\cos A} \right)$

$=(\sec A - \csc A)\left(\dfrac{1 + \sin A \cos A}{\sin A \cos A} \right)$

$=\left(\dfrac{\frac{1}{\cos A} - \frac{1}{\sin A}+\sin A - \cos A}{\sin A\cos A}\right)$

$=\dfrac{\sin A - \sin A \cos^2A - \cos A + \cos A\sin^2A}{(\sin A\cos A)^2}$

$=\dfrac{\sin A(1 - \cos^2A) - \cos A (1 - \sin^2 A)}{(\sin A\cos A)^2}$

$=\dfrac{\sin^3A - \cos^3A}{\sin^2A\cos^2A}$

$=\dfrac{\sin A}{\cos^2A} - \dfrac{\cos A}{\sin^2A}$

$=\left(\dfrac{1}{\cos A}\right)\left(\dfrac{\sin A}{1}\right) - \left(\dfrac{1}{\sin^2A}\right) \left(\dfrac{\cos A}{1}\right)$

$=\sec^2A\csc A - \csc^2A\sec A$

5 0

3 years ago