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Volgvan
3 years ago
12

Could you provide the procedure? Thanks!

Mathematics
1 answer:
Arturiano [62]3 years ago
3 0

i think this is your answer: 12.0 and 12.5

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Line M passes through point (3,1) and is perpendicular to the line of the equation y=3x+4. Which equation describes the graph of
zheka24 [161]

Answer:

y = -1/3x + 2

Step-by-step explanation:

The gradient of the given line is 3 because (y = mx +c where m is the gradient)

Therefore, to find the gradient of the perpendicular line (at 90 degrees), you need to find the negative reciprocal.

The negative reciprocal of 3 is -1/3 because imagine if 3 = 3/1, to get the reciprocal, you flip it, and to get the negative, you just flip the sign.

Now we know that Line M is y = -1/3x + c, we need to find the y-intercept.

To do this, just input the point (3,1) into y = -1/3x + c, to get c. This is because we know (3,1) is on the line from the question.

So it would be 1 = (-1/3 x 3) +c

Which would be 1 = -1 +c

And so c = 2

Put everything together and you get y = -1/3x + 2

8 0
3 years ago
Can someone please help?
Pepsi [2]

Answer:

Try working from Right to left

Step-by-step explanation:

That way the exponits will work out.

6 0
3 years ago
Which of the following best describes the equation below?<br><br> y = |x| + 5
sleet_krkn [62]

Answer:

On a graph, you go up 5.

4 0
3 years ago
What is the area of the figure:
Luba_88 [7]

For this case we have that the area of the triangle is given by:

A = \frac {b * h} {2}

Where:

b: It's the base

h: It's the height

We have to:

cos (45) = \frac {b} {24}\\b = 24 * cos (45)\\b = \frac {\sqrt {2}} {2} * 24\\b = 12 \sqrt {2}

The atura will be given by:

sin (45) = \frac {h} {24}\\h = 24 * sin (45)\\h = \frac {\sqrt {2}} {2} * 24\\h = 12 \sqrt {2}

So, the area is:

A = \frac {12 \sqrt {2} * 12 \sqrt {2}} {2}\\A=\frac{(12\sqrt{2})^2}{2}\\A = 144

Answer:

144

6 0
3 years ago
Read 2 more answers
Deborah has $56 in her bank account currently. She earns $8 per hour at her job at the local ice cream shop. Micah has $32 in hi
Scilla [17]
5 hours I did this on Plato
4 0
3 years ago
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