All categories

3 years ago

The area of a trapezoid is 1400 cm and the bases measure 20 cm and 50 cm, find the height.

Mathematics

1 answer:

arlik [135]3 years ago

4 0

Using the formulaA=a+b2hSolving forhh=2Aa+b=2·1400 20+50=40cm

You might be interested in

According to a census performed in a certain country in a particular year, about 209,280,000 of it's residents were adults, and

gayaneshka [121]

54160000/209280000

= 0.2587
=25.88%

7 0

3 years ago

Evaluate 2x2y3 when x=2\3 and y=1\2

Evgesh-ka [11]

Answer:

The answer is 4

Step-by-step explanation:

2(2/3)2(1/2)3

5 0

3 years ago

PLEASE HELP, THANK YOU! :)

elena-s [515]

Steps to solve:

-4z + 4 + 12z = 3z + 4 + 5z

~Combine like terms

8z + 4 = 8z + 4

~Subtract 4 to both sides

8z = 8z

~Divide 8 to both sides

0 = 0

All real numbers are solutions.

Set builder notation: {x | R}

This is an identity since every value can be a solution to the equation.

Best of Luck!

8 0

3 years ago

Please help me! Find the number x such that f(x) =1

STatiana [176]

Answer:

Step-by-step explanation:

We have the piecewise function:

$f(x) = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

And we want to find x such that f(x)=1.

So, let's substitute 1 for f(x):

$1 = \left\{ \begin{array}{ll} -\frac{1}{2}x-1 & \quad x \leq -2 \\ x & \quad x > -2 \end{array} \right.$

This has two equations. So, we can separate them into two separate cases. Namely:

$1=-\frac{1}{2}x-1\text{ or } 1=x$

Let's solve for x in each case.

Case I:

We have:

$1=-\frac{1}{2}x-1$

Add 1 to both sides:

$2=-\frac{1}{2}x$

Let's cancel out the fraction by multiplying both sides by -2. So:

$2(-2)=(-2)\frac{-1}{2}x$

The right side cancels:

$-4=x\\$

Flip:

$x=-4$

So, x is -4.

Case II:

We have:

$1=x$

Flip:

$x=1$

This is the solution for our second case.

So, we have:

$x_1=-4\text{ or } x_2=1$

Now, can check to see if we have to to remove solution(s) that don't work.

Note that x=-4 is the solution to our first equation.

The first equation is defined only if x is less than -2.

-4 is less than -2. So, x=-4 is indeed a solution.

x=1 is the solution to our second equation.

The second equation is defined only if x is greater than or equal to -2.

1 is greater than or equal to -2. So, x=1 is also a solution.

Therefore, our two solutions are:

$x_1=-4\text{ or } x_2=1$

Out of our answer choices, we can pick D.

And we're done!

7 0

3 years ago

G es to prove that if a, b, and c are integers, where a ≠ 0 and c ≠ 0, such that ac | bc, then a |

Alex Ar [27]

$ac\mid bc$ means there exists an integer $n$ such that $ac=bcn$ . We can divide both sides by $c$ because $c\neq0$ . This implies $a=bn$ for some integer, which is identically $a\mid b$ .

3 0

4 years ago