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3 years ago

Halfway between -20 and 10

Mathematics

1 answer:

VMariaS [17]3 years ago

3 0

Answer:

$-5$

Step-by-step explanation:

Add the two numbers and divide by 2, $-20 + 10 = -10$ , $-10 /2 = -5$

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The equation a=1/2(b^1+b^2)h can be determined the area, a, of a trapezoid with height, h, and base lengths, b^1 and b^2 Which a

Evgesh-ka [11]

The complete question is as follows.

The equation a = $\frac{1}{2}(b_1 + b_2 )h$ can be used to determine the area , a, of a trapezoid with height , h, and base lengths, $b_1$ and $b_2$ . Which are equivalent equations?

(a) $\frac{2a}{h} - b_2 = b_1$

(b) $\frac{a}{2h} - b_2 = b_1$

(d) $\frac{2a}{b_1 + b_2} = h$

(e) $\frac{a}{2(b_1 + b_2)}$ = h

Answer: (a) $\frac{2a}{h} - b_2 = b_1$ ; (d) $\frac{2a}{b_1 + b_2} = h$ ;

Step-by-step explanation: To determine $b_1$ :

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = ( $b_1 + b_2$ )h

$\frac{2a}{h} = b_1 + b_2$

$\frac{2a}{h} - b_2 = b_1$

To determine h:

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = $(b_1 + b_2)h$

$\frac{2a}{(b_1 + b_2)}$ = h

To determine $b_2$

a = $\frac{1}{2}(b_1 + b_2 )h$

2a = $(b_1 + b_2)h$

$\frac{2a}{h} = (b_1 + b_2)$

$\frac{2a}{h} - b_1 = b_2$

Checking the alternatives, you have that $\frac{2a}{h} - b_2 = b_1$ and $\frac{2a}{(b_1 + b_2)}$ = h, so alternatives A and D are correct.

4 0

4 years ago

What is the solution x+2y=7 and x-2y=-1

erica [24]

$\underline{+\left\{\begin{array}{ccc}x+2y=7\\x-2y=-1\end{array}\right}\qquad\text{add both sides of the equations}\\\\.\qquad\qquad2x=6\qquad\text{divide both sides by 2}\\.\qquad\qquad \boxed{x=3}\\\\\text{Put the value of x to the first equation}\\\\3+2y=7\qquad\text{subtract 3 from both sides}\\\\2y=4\qquad\text{divide both sides by 2}\\\\\boxed{y=2}\\\\Answer:\ \boxed{x=3\ and\ y=2\to(3,\ 2)}$