PLEASE HELP me with this!! no explanation needed:)

Pls answer asap :)

dezoksy [38]

Answer:

one fourth

Step-by-step explanation:

8 0

3 years ago

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The circle passes through the point (-1,-6). What is its radius?

love history [14]

Answer:

the radius is something in math

3 0

3 years ago

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A total of 50 juniors and seniors were given a mathematics test. The 35 juniors attained an average score of 80 while the 15 sen

Alik [6]

Answer:

77

Step-by-step explanation:

Firstly, we need to calculate the total score of the junior students and the total score of the senior students.

The total score of the junior students is 35 * 80 = 2,800

The total score of the senior students is 15 * 70 = 1050

The total score is thus 2,800 + 1,050 = 3,850

The average score of the 50 students is thus 3,850/50 which equals 77

8 0

3 years ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

7 0

3 years ago

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What is the solution to the quadratic equation 8m^2+7m-15=-7 ?

Aleks04 [339]

Make equal to zero
add7 both sides
8m^2+7m-8=0
quadratic fomrula

if you have
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$

8m^2+7m-8=0
a=8
b=7
c=-8

x= $\frac{-7+/- \sqrt{7^{2}-4(8)(-8)} }{2(8)}$
x= $\frac{-7+/- \sqrt{49+256} }{16}$
x= $\frac{-7+/- \sqrt{305} }{16}$

x= $\frac{-7+ \sqrt{305} }{16}$ or x= $\frac{-7- \sqrt{305} }{16}$

aprox
x=-1.52902 or 0.654016

4 0

3 years ago