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3 years ago

What 6 and 7 find Missing number 56 and

Mathematics

1 answer:

stiv31 [10]3 years ago

3 0

6 x 7 = 42

And the missing number is 8, because 7 x 8 = 56.

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Find the<br> area of the figure.

EastWind [94]

Answer:

22 sq ft

Step-by-step explanation:

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3 years ago

What are the period and amplitude of the function?

umka2103 [35]

Answer:

The period of function is 5 and amplitude is 3.

Step-by-step explanation:

<em>The period of the function is the time interval after which it repeats itself.</em>

It means the time interval after which the same value of graph is attained.

In the above case looking at the peak point at time 5 seconds, the same value is obtained at time 10 seconds.

Thus the time interval is (10 - 5) 5 seconds.

<em>Amplitude is the maximum value attained by the graph in the complete interval.</em>

In the above case, the maximum value attained is 3.

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3 years ago

Simplify sin^2y/sec^2 y−1 to a single trigonometric function

liubo4ka [24]

Answer:

$\frac{ { \sin}^{2} y}{ { \sec}^{2}y - 1 } = { \cos }^{2} y$

Step-by-step explanation:

We know that ${ \tan }^{2} y = { \sec }^{2} y - 1$

Also , ${ \tan}^{2} y = \frac{ { \sin }^{2} y}{ { \cos }^{2}y }$

So ,

$\frac{ { \sin }^{2}y }{ { \sec }^{2}y - 1 } = \frac{ { \sin}^{2} y}{ { \tan }^{2} y} = \frac{ { \sin }^{2}y }{ \frac{ { \sin}^{2} y}{ { \cos}^{2}y } } = { \cos }^{2} y$

6 0

2 years ago

What is the solution to the equation −p − 13 = −4 /3 p + 2/3 ( 5p − 6 ) ?

jekas [21]

Answer:

The answer is p= 2 and 1/3

Step-by-step explanation:

This answer can be found using the order of operations to solve for p.

Start by distributing 2/3 by (5p-6)

add -4/3p to 10/3p equaling 6/3p or 2p

subtract 13 from both sides

-p= 2p - 7

p=2 and 1/3

3 0

2 years ago

100 points! Please help asap! Look at the picture attached. My teacher said that both 1 and 2 are neither can someone explain wh

Neko [114]

Answer:

Your teacher is right, there is not enough info

Step-by-step explanation:

<h3>Question 1</h3>

We can see that RS is divided by half

The PQ is not indicated as perpendicular to RS or RQ is not indicates same as QS

So P is not on the perpendicular bisector of RS

<h3>Question 2</h3>

We can see that PD⊥DE and PF⊥FE

There is no indication that PD = PF or ∠DEP ≅ FEP

So PE is not the angle bisector of ∠DEF

6 0

3 years ago