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3 years ago

Need an answer to this (all help appreciated)

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

5 0

Answer:

My answer is second one

Step-by-step explanation:

Also you didn’t give me enough information, so it’s a guess.

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7x + 3y = m

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7x + 3y = m

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Step-by-step explanation:

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2 years ago

Find the exponential function that satisfies the given conditions: Initial value = 70, decreasing at a rate of 0.5% per week

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Y = 70 e^(.005x) where x is the time in weeks

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The lines represented by the equations y = -X – 3 and

xz_007 [3.2K]

The equations y = -x -3 and 5y + 5x = -15 represents the same line option 'the same line' is correct.

<h3>What is linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have two equation of the line:

y = -x -3 and

5y + 5x = -15

From the equation 5y + 5x = -15:

Divide by 5 on the above equation:

y + x = -3

y = -x -3

The two equations y = -x -3 represents the same line.

Thus, the equations y = -x -3 and 5y + 5x = -15 represents the same line option 'the same line' is correct.

Learn more about the linear equation here:

brainly.com/question/11897796

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1 year ago

Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and

zzz [600]

Answer:

$\bold{sin(38^\circ)=cos(52^\circ)}$

Step-by-step explanation:

Given that $\triangle KJL$ is a right angled triangle.

$\angle JKL = 52^\circ\\\angle KLJ = 38^\circ$

and

$\angle KJL = 90^\circ$

Kindly refer to the attached image of $\triangle KJL$ in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given $\triangle KJL$ .

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}$

$sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}$ ....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

$cos\theta =\dfrac{Base}{Hypotenuse}$

$cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}$ ....... (2)

By (1) and (2):

sin(38°) $\neq$ cos(38°).

$cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}$ ...... (3)

Comparing equations (1) and (3):

we get the both are same.

$\therefore \bold{sin(38^\circ)=cos(52^\circ)}$

6 0

3 years ago