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alex41 [277]
2 years ago
5

Pleaseeee helpppppppppp

Mathematics
1 answer:
Georgia [21]2 years ago
3 0

Answer:

A

Step-by-step explanation:

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0.4/100 simplified if it can be
zysi [14]

Answer:

vjneovfvfvvfijbnvnbwhevwdjnobfhijnvow

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Given ab and cb are tangents of p, and m =10°. What is the measure of abp?
yanalaym [24]

Answer:

Hence, ∠ABP=10°

Step-by-step explanation:

We are given that:

AB and CD are tangents of the circle.

Now we are given that m∠ACP=10°.

Also we are given that AP and CP are radius of the given circle that means length of AP and CP are equal.

Also we know that angle opposite to the equal sides are equal that means:

∠ACP=∠PAC.

Hence , ∠PAC=10°.

Also in a triangle the sum of all the angles is 180°.

i.e. ∠ACP+∠PAC+∠APC=180°

this means that 10+10+∠APC=180°

∠APC=180-20

∠APC=180-20=160°

As AB and CD are tangents to the circles that means:

∠BAP=∠BCP=90°

Also APCB is a quadrilateral and we know that sum of all the angles of a quadrilateral=360°

Hence,

∠BAP+∠BCP+∠APC+∠ABC=360°

⇒ 90+90+160+∠ABC=360

⇒ 180+160+∠ABC=360

⇒ 340+∠ABC=360

⇒ ∠ABC=360-340=20°

also ∠ABP=(1/2)∠ABC

Hence, ∠ABP=(1/2)×20°=10°

Hence, ∠ABP=10°

7 0
3 years ago
Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)
vfiekz [6]

Answer:

Step-by-step explanation:

given a point (x_0,y_0) the equation of a line with slope m that passes through the  given point is

y-y_0 = m(x-x_0) or equivalently

y = mx+(y_0-mx_0).

Recall that a line of the form y=mx+b, the y intercept is b and the x intercept is \frac{-b}{m}.

So, in our case, the y intercept is (y_0-mx_0) and the x  intercept is \frac{mx_0-y_0}{m}.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph (x_0,\frac{1}{x_0}). Which means that y_0=\frac{1}{x_0}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

y' = \frac{-1}{x^2}. So evaluated at x_0 we get m = \frac{-1}{x_0^2}

Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0
3 years ago
Write10:15 pm in 24 hour time
Stels [109]
Your answer is 22:15
5 0
3 years ago
Please help with solving
Makovka662 [10]
N.O = 4
N is midpoint of M.0
Meaning M.N also has to be 4
4+4= 8
N.P = 6
0.P = 2
8+ 2 = 10

4 0
2 years ago
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