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2 years ago

What is the area of a quadrilateral with vertices of (2,-4) (4,-4) (2,1) and (4,1)

Mathematics

1 answer:

taurus [48]2 years ago

4 0

if you draw it out, it is a vertical rectangle

height of rectangle= 1-(-4)= 5units

length of rectangle= 4-2= 2units

area of rectangle =2×5= 10units^2

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Can someone answer this for me pls and thank you

kherson [118]

x in terms of b:

-2(bx - 5) = 16

-2bx + 10 = 16

-2bx = 6

x = 6 / -2b

x = -3/b

--------------------------------

when b = 3 then

-2(3x - 5) = 16

-6x + 10 = 16

-6x = 6

x = -1

Answer

The value of x in terms of b is -3/b

The value of x when b is 3 is -1

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2 years ago

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enyata [817]

ANSWER:
true

STEP-BY-STEP EXPLANATION:

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Five friends plan to split a restaurant bill evenly. The total cost of the meal is $89.75, and they want to leave a 20% tip. Wha

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2 years ago

Find the value of x in the triangle shown below

CaHeK987 [17]

Answer:

Step-by-step explanation:

Apply sine formula:

$\frac{sin \: a}{a} = \frac{sin \: b}{b} = \frac{sin \: c}{c}$

$\frac{sin \: a}{a} = \frac{sin \: c}{c}$

Plug the values

$\frac{sin \: (88)}{6.9} = \frac{sin \: x}{5}$

Apply cross product property

$5 \: sin \: (88) = 6.9 \: (sin \: x)$

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$x = {sin}^{ - 1} ( \frac{5 \: sin \:( 88)}{6.9} )$

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Hope this helps .....

Best regards!!!

3 0

3 years ago

Read 2 more answers

5. Solve using any method.<br> y = -x + 8<br> y=x+4

vlada-n [284]

Answer:

Step-by-step explanation:

$\bold{ELIMINATION\ METHOD}\\\\\left\{\begin{array}{ccc}y=-x+8\\y=x+4&\TEXT{change the signs}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}y=-x+8\\-y=-x-4\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=-2x+4\qquad\text{add}\ 2x\ \text{to both sides}\\.\qquad2x=4\qquad\text{divide both sides by 2}\\.\qquad\boxed{x=2}\\\\\text{Put it to the second equation}\\y=2+4\\\boxed{y=6}$

$\bold{SUBSTITUTION\ METHOD}\\\\\left\{\begin{array}{ccc}y=-x+8&(1)\\y=x+4&(2)\end{array}\right\\\\\text{Substitute (1) to (2)}\\\\-x+8=x+4\qquad\text{subtract 8 from both sides}\\-x=x-4\qquad\text{subtract}\ x\ \text{from both sides}\\-x-x=-4\\-2x=-4\qquad\text{divide both sides by (-2)}\\\boxed{x=2}\\\\\text{Put it to the second equation}\\y=2+4\\\boxed{y=6}$

7 0

2 years ago