A=1/2h(B+b); a=6,b=2, b=1 What is h

Mathematics

1 answer:

alexdok [17]2 years ago

3 0

A=(1/2)(h)(B+b)
(6)=(1/2)(h)((2)+(1))
12=h(3)
h=4

Answer: h=4

You might be interested in

The equation x+53=15 can be solved by performing two operations on both sides.

Wewaii [24]

Hey!

You're right! This equation can definitely be solved by performing two operations on both sides. Let me show you how it's done.

First, let's write out our equation.

Original Equation :
x + 53 = 15

Now that that's done, we'll be performing two operations on both sides. The operation we'll want to do is subtracting both sides of the equation by 53. We do this to get x on its own.

Original Equation :
x + 53 = 15

New Equation {Added Subtract 53 to Both Sides} :
x + 53 - 53 = 15 - 53

Now we have to solve the equation. Let's do the left side first.

Left Side of the Equation :
x + 53 - 53

Left Side of the Equation {Solved} :
x

Now, we'll solve the right side of the equation.

Right Side of the Equation :
15 - 53

Right Side of the Equation {Solved} :
-38

Now we can put both solutions to both sides of the equation together.

New Equation :
x = -38

Since this cannot be simplified any farther, this is our final answer. And that's it!

So, now we know that in the equation x + 53 = 15, x = -38

Hope this helps!

- Lindsey Frazier ♥

3 0

2 years ago

A person plans to drive across country 1,798 miles. At what speed would they have to drive in order to drive no more than 30 hou

pychu [463]

Answer:

59.9 mph or rounded to the nearest tenth (60 mph)

Step-by-step explanation:

Speed=Distance÷time

Distance-1798

Time-30

Speed-?

1798÷30=59.9mph or 60mph

6 0

1 year ago

Read 2 more answers

(x + 1)^2 = 3x – 1 solving Quadratics ?

Deffense [45]

Step-by-step explanation:

$(x + 1)^{2} = 3x - 1 \\ \\ \therefore \: {x }^{2} + 2x + 1 = 3x - 1 \\ \\ \therefore \: {x }^{2} + 2x + 1 - 3x + 1 = 0\\ \\ \therefore \: {x }^{2} + 2x - 3x + 1 - 1 = 0 \\ \\ \therefore \: {x }^{2} - x = 0 \\ \\ \therefore \: x(x - 1) = 0 \\ \\ \therefore \: x = 0 \: \: or \: \: x - 1 = 0 \\ \\ \therefore \: x = 0 \: \: or \: \: x = 1 \\ \\ \huge{ \red{ \boxed{\therefore \: x = \{0, \: \: 1 \}}}}$