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3 years ago

Help me pls need helppppppp

Mathematics

2 answers:

seraphim [82]3 years ago

7 0

Answer:

9 I think I'm not entirely sure

bonufazy [111]3 years ago

6 0

The answer is 11 bc it does have “b” next to it

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Points P, Q, R and S lie on a line in that order. Q is the midpoint of PR. If the length of RS is 9 and the length of PS is 17,

Marta_Voda [28]

The length of PQ is 10

4 0

4 years ago

Line segment GT contains the point G(−3, 5) and a midpoint at A(1, −4). What is the location of endpoint T?

algol [13]

Imagine you're moving along the segment. Since the midpoint is in the middle of the segment (obviously), it means that when you've traveled from G to A, you're halfway through your journey, along both x and y directions. So, let's break the problem in two and analyze both directions.

Along the x axis, you've moved from -3 to 1, so you moved 4 units forward. This means that you have 4 units still to go, and your journey will end at coordinate 5.

Similarly, along the y axis, you've moved from 5 to -4, so you moved 9 units downward. This means that you have 9 units still to go, and your journey will end at coordinate -13.

So, the coordinates of the endpoint are $T = (5,-13)$

If you prefer a more analyitical approach, simply write the definition of the midpoint and solve it for the coordinates of T.

We have $G = (-3, 5)$ and $T = (x_T,y_T)$ . The midpoint is computed as

$A = \left( \frac{-3+x_T}{2},\frac{5+y_T}{2} \right) = (1, -4)$

So, you have the equations

$\frac{-3+x_T}{2} = 1,\qquad \frac{5+y_T}{2} = -4$

Multply both equations by 2 to get

$-3+x_T = 2,\qquad 5+y_T = -8$

Move the constants to the right hand sides to get

$x_T = 5,\qquad y_T = -13$

8 0

3 years ago

Solve the above que no. 55

aleksandr82 [10.1K]

Answer:

Let $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ , we proceed to prove the trigonometric expression by trigonometric identity:

1) $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ Given

2) $\left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)$ $\tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}$

3) $\left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)$

4) $\left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)$ $\sin^{2}A+\cos^{2}A = 1$

5) $\frac{1}{\sin^{2}A\cdot \cos^{2}A}$

6) $\frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}$ $\sin^{2}A+\cos^{2}A = 1$

7) $\frac{1}{\sin^{2}A-\sin^{4}A}$ Result

Step-by-step explanation:

Let $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ , we proceed to prove the trigonometric expression by trigonometric identity:

1) $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ Given

2) $\left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)$ $\tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}$

3) $\left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)$

4) $\left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)$ $\sin^{2}A+\cos^{2}A = 1$

5) $\frac{1}{\sin^{2}A\cdot \cos^{2}A}$

6) $\frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}$ $\sin^{2}A+\cos^{2}A = 1$

7) $\frac{1}{\sin^{2}A-\sin^{4}A}$ Result

4 0

3 years ago

You cut 63 meters of rope into 9 equal pieces. How long is each equal piece of rope?

Rama09 [41]

7 Meters. 63/9=7

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It's too short.<span> Write at least 20 characters to explain it well.</span>

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4 years ago

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1) Find the mean for the given sample data. Unless indicated otherwise, round your answer to one more decimal place than is pres

Semmy [17]

Alright, so #1 is C, 4.88.

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4 years ago

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