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Anika [276]
3 years ago
9

Solve the right triangle. Round decimal answers to the nearest tenth. E D 8 14 1 F DE ~ ,mZE ~ MZD​

Mathematics
1 answer:
Ugo [173]3 years ago
6 0

Answer:

DE≈16.1

m<E≈60.3

m<D≈29.7

Step-by-step explanation:

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Select the equivalent expression.
ZanzabumX [31]

the equivalent expression is C

3 0
3 years ago
Help help help help help
Mkey [24]

Answer:

  x = 42

Step-by-step explanation:

The marked angles are supplementary, so their sum is 180°.

  (2x +8) +(2x +4) = 180

  4x +12 = 180 . . . . . . . . . simplify

  x +3 = 45 . . . . . . . divide by 4 (because we can)

  x = 42 . . . . . . subtract 3

_____

<em>Additional comment</em>

A "two-step" linear equation like this one is usually solved by subtracting the unwanted constant, then dividing by the coefficient of the variable. Here, we have done those steps in reverse order. This makes the numbers smaller and  eliminates the coefficient of the variable. Sometimes I find it easier to solve the equation this way.

5 0
3 years ago
Which of the following is the graph of y=3 sec [ 2(x-pi/2)] + 2
asambeis [7]

Answer:

A.) The first graph

Step-by-step explanation:

3 0
3 years ago
What is the length of the conjugate axis? (y-2)^2/16 -(x+1)^2/144=1
aleksandr82 [10.1K]

Answer: 24

A hyperbola is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point in the same plane to its distance from a fixed line is always constant, which is always greater than unity.

the fixed point is called the focus and the fixed line is directrix and the ratio is the eccentricity.

The general equation for the vertical hyperbola is

            [ (y-k)^2 / a^2 ] – [ (x-h)^2 / b^2 ] = 1

The conjugate axis of the vertical hyperbola is y = k

Length of the conjugate axis  = 2b

According to the question k = 2, h = -1, a = 4, b = 12

Length of the conjugate axis = 2b = 2 * 12  = 24

Learn more about hyperbola here :

brainly.com/question/3351710

#SPJ9

7 0
2 years ago
Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)
Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

x^{2} +y^{2}+4x-1=0

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

(x-h)^2+(y-k)^2=r^2

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

      (\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2}  )

    (-2,0)

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}

r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}

r=\sqrt{5}

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius r=\sqrt{5}

The standard form of circle equation

(x-h)^2+(y-k)^2=r^2

(x-(-2))^2+(y-0)^2=\sqrt{5} ^2

on simplification is

x^{2} +y^{2}+4 x-1=0

5 0
3 years ago
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