I think eight because if you add those days together they will play again in eight days so that they play on the same day.

Sorry if I got it wrong I tried and that's what counts, right?

3 0

2 years ago

What is the solution to the system of equations? Use the substitution method. {y=2x+43x−6y=3

garri49 [273]

We have that
y=2x+4--------> equation 1
3x−6y=3-------> equation 2

step 1
I substitute the value of y in equation 1 for the value of y in equation 2
so
3x−6*[2x+4]=3-------> 3x-12x-24=3
-9x=3+24
-9x=27------> 9x=-27
x=-27/9
x=-3

step 2
I substitute the value of x in equation 1 to get the value of y
y=2x+4--------> y=2*(-3)+4--------> y=-6+4
y=-2

the answer is
the solution is the point (-3,-2)
x=-3
y=-2

5 0

3 years ago

Solve. 9x+9> 50.4 Enter the answer in the box. X >

coldgirl [10]

Rearrange unknown terms to the left side of the equation

$\boldsymbol{\sf{ 9x > 50.4-9 \ \ \longmapsto \ \ \ [Subtract]}}$

Calculate the sum or difference.

$\boldsymbol{\sf{9x > 41.4 }}$

Convert decimal to fraction.

$\boldsymbol{\sf{9x > \dfrac{414}{10} }}$

Reduce the greatest common factor for both sides of the inequality.

$\boldsymbol{\sf{x > \dfrac{46}{10} }}$

Reduce the fraction

$\red{ \boxed{\boldsymbol{\sf{\blue{ Answer \ \ \longmapsto \ \ x > \dfrac{23}{5} }}}}}$

3 0

1 year ago

Read 2 more answers

Use the value of the first integral I to evaluate the two given integrals. I = integral^3_0 (x^3 - 4x)dx = 9/4 A. integral^3_0 (

Troyanec [42]

Answer:

A) -9/2

B) 9/4

C) -9/2, same as A)

Step-by-step explanation:

We are given that $I=\int_0^3 x^3-4x dx=9/4$ . We use the properties of integrals to write the new integrals in terms of I.

A) $\int_0^3 8x-2x^3 dx=\int_0^3 -2(x^3-4x) dx=-2\int_0^3 x^3-4x dx=-2I=-9/2$ . We have used that ∫cf dx=c∫f dx.

B) $\int_3^0 4x - x^3 dx=-\int_0^3 (4x-x^3) dx=\int_0^3-(4x-x^3) dx=\int_0^3 x^3-4x dx=I=9/4$ . Here we used that reversing the limits of integration changes the sign of the integral.

C) It's the same integral in A)

4 0

2 years ago