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Airida [17]
3 years ago
15

On monday 2/3 of the team practiced, tuesday 7/8, Wednesday 1/2 and thursday3/4 on which day were the most team members present

at practice
Mathematics
1 answer:
SpyIntel [72]3 years ago
5 0
Monday=2/3=66.7% Tuesday 7/8=87.5% Wednesday=1/2=50% and Thursday=3/4=75% so Tuesday the most team members showed up to practice
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Let A = any rational number is the absolute value of a different if A is a positive number or negative number Explain
artcher [175]

Absolute value is always positive because its talking about the real distance or steps to zero and even when you at a negative number it takes real steps to get back to zero.  

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3 years ago
Express the sum of the polymonial 3x^2+15x-56 and the square of the binomial (x-8) as a polynomial in standard form.
tester [92]

Given:

Polynomial is 3x^2+15x-56.

To find:

The sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form.

Solution:

The sum of given polynomial and the square of the binomial (x-8) is

3x^2+15x-56+(x-8)^2

=3x^2+15x-56+x^2-2(x)(8)+8^2    [\because (a-b)^2=a^2-2ab+b^2]

=3x^2+15x-56+x^2-16x+64

On combining like terms, we get

=(3x^2+x^2)+(15x-16x)+(-56+64)

=4x^2-x+8

Therefore, the sum of given polynomial and the square of the binomial (x-8) as a polynomial in standard form is 4x^2-x+8.

7 0
3 years ago
Which pair of angles is a pair of corresponding?
Agata [3.3K]

Answer:

3 and 8 is a pair of corresponding.

8 0
3 years ago
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To show that TRK= TUK by the SAS postulate, what additional information is necessary?
Anni [7]

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Step-by-step explanation:

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An isosceles trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if p
horrorfan [7]

Answer:

a = -0.3575

Step-by-step explanation:

The points A and D lie on the x-axis, this means that they are the x-intercepts of the parabola, and therefore we can find their location.

The points A and B are located where

y=a(x+1)(x-5)=0

This gives

x=-1

y=5

Now given the coordinates of A, we are in position to find the coordinates of the point B. Point B must have y coordinate of y=2 (because the base of the trapezoid is at y=0), and the x coordinate of B, looking at the figure, must be x coordinate of A plus horizontal distance between A and B, i.e

B_x=A_x+\frac{2}{tan(60)} =-1+\frac{2\sqrt{3} }{3}

Thus the coordinates of B are:

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

Now this point B lies on the parabola, and therefore it must satisfy the equation  y=a(x+1)(x-5).

Thus

2=a((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)

Therefore

a=\frac{2}{((-1+\frac{2\sqrt{3} }{3})+1)((-1+\frac{2\sqrt{3} }{3})-5)}

\boxed{a=-0.3575}

8 0
3 years ago
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