All categories

2 years ago

Help me answer this question :((((

Mathematics

1 answer:

icang [17]2 years ago

8 0

Answer:

97.5

Step-by-step explanation:

75,85,90,105,120,150

90 and 105 are the middle, and median means the middle, but you can only have one answer.

90+105=195

195÷2=97.5

97.5 is the middle of all these numbers

You might be interested in

Please help I'll give brilliantist:)

Vanyuwa [196]

Answer:

(-3, -6)

Step-by-step explanation:

Coordinate = (x-axis, y-axis)

=(x,y)

identify value of coordinate A at x-axis before y-axis

x-axis of A is in the middle of -2 and -4, thus it is -3

y-axis of A when refer to graph is -6

thus answer is (-3,-6)

6 0

3 years ago

.....................

Ivahew [28]

1.) Parallelogram

2.) Quadrilateral

3.) Complimentary

4.) bisect

5.) Supplementary

6.) GH = 14 units

7.) 80°

8.) KF = 10

7 0

2 years ago

What is the value of the expression below when z=6?<br><br> Z to the 2nd power +3z -3

aleksklad [387]

6^2= 36
3 times 6 = 18
36+18-3 =
Answer is 51
I hope this is correct :)

7 0

3 years ago

Given: Circle A externally tangent to Circle B.

Zolol [24]

<h2>Answer:</h2>

B. 3

<h2>Step-by-step explanation:</h2>

A tangent is a line that passes through exactly one point on a circle. For any point on a circle, the radius and tangent line through that point are always perpendicular to each other making a right angle. If two triangles touch each other externally, then there will be 3 lines making right angle with radius of the circles at the point of intersection of circle and tangents. These three tangent lines are shown below. They are the black, blue and green ones.

7 0

3 years ago

Read 2 more answers

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago