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KATRIN_1 [288]
3 years ago
8

The length of the shorter base in an isosceles trapezoid is 4 in, its altitude is 5 in, and the measure of one of its obtuse ang

les is 135°. Find the area of the trapezoid.
Mathematics
1 answer:
il63 [147K]3 years ago
4 0
The answer is 45.

Ask me if you still don't understand it because I didn't write any explanation.
You might be interested in
Is this right I need help with this
Margarita [4]
Y axis is counting by 2's
it started at 2 and went up to 6
Y = 4

X axis is counting by 2'2
starts at one goes to 3
X = 2

Slope is 4/2 because rise over run
4/2 = 2
Slope is simplified to 2. 
5 0
3 years ago
The point (17,−9) is translated to the point (6,2). Select the description of the translation.
zubka84 [21]

Answer:

11 up , 11 left

Step-by-step explanation:

Given

(17, -9) to (6,2)

Required

Determine the translation rule

Considering the x coordinates.

From 17 to 6

6 is to the left of 17.

Using the translation rule:

x = x' + b

17 = 6 + b

17 - 6 = b

11 = b

b = 11

<em>Hence: From 17 to 6 is 11 units left translation</em>

Considering the y coordinates.

From -9 to 2

2 is at the top -9.

Using the translation rule:

y = y' + b

-9 = 2 + b

-9 - 2 = b

-11= b

b = -11

The negative value impies a upward translation.

Hence: From -9 to 2 is 11 units up translation

4 0
3 years ago
PLEASE ANSWER ASAP! YOUR ANSWER MUST INCLUDE AN EXPLANATION IN ORDER TO RECEIVE 10 POINTS AND THE BRAINLIEST ANSWER! THANKS!!!
ZanzabumX [31]
Hello,

Sandy is right .

Other method:
V=168=(10+4)*3*4

3 0
3 years ago
What is the solution to this system of equations?
amm1812

Answer:

Step-by-step explanation:

(i) x+2y=4

(ii) x=4-2y

(iii) 2x-2y=5  

substituting (ii) in (iii)

2x-2y=5

2(4-2y)-2y=5

8-4y-2y=5

-6y=5-8

-6y=-3

(iv) y=\frac{1}{2}              

substituting (iv) in (ii)

x=4-2y

x=4-2×\frac{1}{2}

x=4-1

x=3

∴ B.(3,\frac{1}{2})

HOPE IT HELPS YOU!!!!

3 0
3 years ago
Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [
Darya [45]

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

3 0
3 years ago
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