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2 years ago

Marta got 88 points on her English test. Sandy got x points on the same test. What was their average

Mathematics

2 answers:

Anastasy [175]2 years ago

7 0

Answer: (88+x)/2

Explanation: In order to find average, add all the numbers and divide it by how many numbers you added. So,

(88 + x)/2

Nothing else was given so I could only find this equation.

Brainliest please if I am correct. Hope it helped!

zlopas [31]2 years ago

3 0

Answer:

Their average score was 88 + x / 2.

Step-by-step explanation:

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15 divided by 1/5 is equal to 1.

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1900 people attended a baseball game. 1691 of the people attending supported the home team while, 209 supported the visiting tea

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The answer to the percentage of home team fans in attendance is 89%

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Answer18:

The quadrilateral ABCD is not a parallelogram

Answer19:

The quadrilateral ABCD is a parallelogram

Step-by-step explanation:

For question 18:

Given that vertices of a quadrilateral are A(-4,-1), B(-4,6), C(2,6) and D(2,-4)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-(-1)}{(-4)-(-4)}$

m= $\frac{7}{0}$

The slope is 90 degree

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{6-6}{(-4)-(-1)}$

m= $\frac{0}{(-3)}$

The slope is zero degree

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-4)-6}{2-2}$

m= $\frac{-10}{0}$

The slope is 90 degree

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-(-4)}{(-4)-(2)}$

m= $\frac{3}{-6}$

m= $\frac{-1}{2}$

The slope of the only line AB and CD are the same.

Thus, The quadrilateral ABCD is not a parallelogram

For question 19:

Given that vertices of a quadrilateral are A(-2,3), B(3,2), C(2,-1) and D(-3,0)

The slope of a line is given m= $\frac{Y2-Y1}{X2-X1}$

Now,

The slope of a line AB:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{2-3}{3-(-2)}$

m= $\frac{-1}{5}$

The slope of a line BC:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{(-1)-2}{2-3}$

m= $\frac{-3}{-1}$

m=3

The slope of a line CD:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{0-(-1)}{(-3)-2}$

m= $\frac{-1}{5}$

The slope of a line DA:

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{3-0}{(-2)-(-3)}$

m=3

The slope of the line AB and CD are the same

The slope of the line BC and DA are the same

Thus, The quadrilateral ABCD is a parallelogram

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3 years ago

Points H and F lie on circle c What is the length of line segment GH?

Nina [5.8K]

Answer:

6 units

Step-by-step explanation:

Given: Points H and F lie on circle with center C. EG = 12, EC = 9 and ∠GEC = 90°.

To find: Length of GH.

Sol: EC = CH = 9 (Radius of the same circle are equal)

Now, GC = GH + CH

GC = GH + 9

Now In ΔEGC, using pythagoras theorem,

$(Hypotenuse)^{2} = (Base)^{2} +(Altitude)^{2}$ ......(ΔEGC is a right triangle)

$(GC)^{2} = (GE)^{2} +(EC)^{2}$

$(GH + 9)^{2} = (9)^{2} +(12)^{2}$

$(GH )^{2} + (9)^{2} + 18GH = 81 + 144$

$(GH )^{2} + 81 + 18GH = 81 + 144$

$(GH )^{2} + 18GH = 144$

Now, Let GH = <em>x</em>

$x^{2} +18x = 144$

On rearranging,

$x^{2} +18 x - 144 = 0$

$x^{2} - 6x +24x + 144 = 0$

$x (x-6) + 24 (x - 6) =0$

$(x - 6) (x + 24) = 0$

So x = 6 and x = - 24

∵ x cannot be - 24 as it will not satisfy the property of right triangle.

Therefore, the length of line segment GH = 6 units. so, Option (D) is the correct answer.

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Step-by-step explanation:

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