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Lelechka [254]
3 years ago
5

Find the slope intercept form -2/5y = -2

Mathematics
1 answer:
melomori [17]3 years ago
3 0

Answer:

y =  5

Step-by-step explanation:

-2y/5 = -2

(Multiply both sides by 5/-2)

y =  5

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Patricia annual Salary was 52,000. She earned a 6% raise what is her new salary
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2 square5 + 5 square5
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Answer:

=2^5+5^5

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3 years ago
DIRECTIONS: Road the question and select the best respons
lana66690 [7]

Answer:

C. 510 cm^2

Step-by-step explanation:

Well to find TSA or Total Surface Area,

We need to find the area of al the triangles and rectangles.

Let's start with the 2 rectangles facing forwards.

They both have dimensions of 5*15 and 12*15,

75 + 180

= 255 cm ^2

Now let's do the back rectangle which has dimensions of 15 and 13.

15*13 = 195 cm^2

Now we can do the top and bottom triangles,

Since we don't have height we can use the following formula,

A = \sqrt{S(S-a)(S-b)(S-c)}

S is S = \frac{1}{2} (A+B+C)

S= 15

Now with s we can plug that in,

A = \sqrt{15(15-5)(15-13)(15-12)}

The a b and c are the sides of the triangle.

So let's solve,

15 - 5 = 10

15 - 13 = 2

15 - 12 = 3

10*2*3 = 60

60*15 = 900

\sqrt{900} = 30 cm^2

Since there is 2 triangles with the same dimensions their areas combined is 60 cm^2

60 + 255 + 195 = <u><em>510 cm^2</em></u>

<em>Thus,</em>

<em>the TSA of the right triangular prism is C. 510 cm^2.</em>

<em>Hope this helps :)</em>

7 0
3 years ago
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The line l is tangent to the circle with equation x^2 + y^2=10 at the point P.
babunello [35]

Given:

The equation of a circle is

x^2+y^2=10

A tangent line l to the circle touches the circle at point P(1,3).

To find:

The equation of the line l.

Solution:

Slope formula: If a line passes through two points, then the slope of the line is

m=\dfrac{y_2-y_1}{x_2-x_1}

Endpoints of the radius are O(0,0) and P(1,3). So, the slope of radius is

m_1=\dfrac{3-0}{1-0}

m_1=\dfrac{3}{1}

m=3

We know that the radius of a circle is always perpendicular to the tangent at the point of tangency.

Product of slopes of two perpendicular lines is always -1.

Let the slope of tangent line l is m. Then, the product of slopes of line l and radius is -1.

m\times m_1=-1

m\times 3=-1

m=-\dfrac{1}{3}

The slope of line l is -\dfrac{1}{3} and it passs through the point P(1,3). So, the equation of line l is

y-y_1=m(x-x_1)

y-3=-\dfrac{1}{3}(x-1)

y-3=-\dfrac{1}{3}(x)+\dfrac{1}{3}

Adding 3 on both sides, we get

y=-\dfrac{1}{3}x+\dfrac{1}{3}+3

y=-\dfrac{1}{3}x+\dfrac{1+9}{3}

y=-\dfrac{1}{3}x+\dfrac{10}{3}

Therefore, the equation of line l is y=-\dfrac{1}{3}x+\dfrac{10}{3}.

4 0
2 years ago
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