On the weekend, I spent 6 3/5 hours reading and 1 1/4 hours at the gym. How much less time did I spend at the gym then reading?

Solve each quadratic equation. Show your work. 14. (2x – 1)(x + 7) = 0 15. x2 + 3x = 10 16. 4x2 = 100

14. (2x - 1)(x + 7) = 0

Using the zero factor property, we know that either the first or second terms (or both) must be equal to 0 if their product is 0. We can set each term equal to 0 to find the solutions:

2x - 1 = 0
2x = 1
x = 1/2

x + 7 = 0
x = -7

15. $x^{2} +3x=10$

To solve this equation, you first need to set it equal to 0:

$x^{2} +3x=10 \\ x^{2} +3x-10 = 0$

Next, it can be factored:

$x^{2} +3x-10 = 0 \\ (x+5)(x-2)=0$

Finally, we can solve just like we did above:

x + 5 = 0
x = -5

x - 2 = 0
x = 2

16. $4x^2=100$

First, you can simplify by dividing each side by 4:

$4x^2=100 \\ x^2=25$

Now, set the equation equal to 0:

$x^2=25 \\ x^2-25=0$

Next, factor:

$x^2-25=0 \\ (x+5)(x-5)=0$

Finally, find the solutions:

x + 5 = 0
x = -5

x - 5 = 0
x = 5

6 0

3 years ago

Pls Help! John wants to prove ABC = to DEF. He Knoes AB = DE and AC = DF. With Screenshot

matrenka [14]

Answer:

$\angle A \cong \angle D$

Step-by-step explanation:

The Side-Angle-Side method cana only be used when information given shows that an included angle which is between two sides of a ∆, as well as the two sides of the ∆ are congruent to the included side and two sides of the other ∆.

Thus, since John already knows that $\overline{AB} \cong \overline{DE}$ and $\overline{AC} \cong \overline{DF}$ , therefore, an additional information showing that the angle between $\overline{AB}$ and $\overline{DE}$ in ∆ABC is congruent to the angle between $\overline{AC}$ and $\overline{DF}$ in ∆DEF.

For John to prove that ∆ABC is congruent to ∆DEF using the Side-Angle-Side method, the additional information needed would be $\angle A \cong \angle D$ .

See attachment for the diagram that has been drawn with the necessary information needed for John to prove that ∆ABC is congruent to ∆DEF.