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Andru [333]
3 years ago
8

In the diagram below, OPis circumscribed about quadrilateral ABCD. What is the value of x?

Mathematics
1 answer:
Nikitich [7]3 years ago
7 0

9514 1404 393

Answer:

  C.  82

Step-by-step explanation:

Opposite angles of an inscribed quadrilateral are supplementary. So ...

  (x + 10°) + 88° = 180°

  x = 180° -98°

  x = 82°

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Change 1/5 to a decimal with work shown
uranmaximum [27]

Answer:

.2

1/5=2/10 2/10=.2

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3 years ago
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What is 25m+100-24m-75=68<br> Please explain
Mars2501 [29]

Let's solve your equation step-by-step.

25m+100−24m−75=68

Step 1: Simplify both sides of the equation.

25m+100−24m−75=68

25m+100+−24m+−75=68

(25m+−24m)+(100+−75)=68(Combine Like Terms)

m+25=68

m+25=68

Step 2: Subtract 25 from both sides.

m+25−25=68−25

m=43

Answer:

m=43

8 0
4 years ago
4. Which of the quadratic functions has the narrowest graph?
iris [78.8K]

Answer:

y=2x^2

Step-by-step explanation:

when you apply a number greater than 1 to the x value, it compresses the x value by the denominator of the fraction.

4 0
3 years ago
Does this have infinite solutions or no solutions?
Papessa [141]

Answer:

when they cross u should get a perfect solution of (2,6)

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3 years ago
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Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

7 0
3 years ago
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