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Veseljchak [2.6K]
3 years ago
11

Trapezoid WKLX has vertices W(2, −3), K(4, −3), L(5, −2) , and X(1, −2) . Trapezoid WKLX ​ is translated 4 units right and 3 uni

ts down to produce trapezoid trapezoid W'K'L'X' .
Which coordinates describe the vertices of the image?

W'(6, 0), K'(8, 0), L'(9, 1) , and X'(5, 1)
W′(6, −6), K′(8, −6), L′(9, −5) , and X′(5, −5)
W'(5, 1), K'(7, 1), L'(8, 2) , and X′(4, 2)
W′(−1, 1), K′(1, 1), L′(2, 2) , and X′(−2, 2)
Mathematics
2 answers:
guapka [62]3 years ago
7 0
Okay, so first of all, if the trapezoid was translated right 4,
Then all the (x)'s should be 4 more than the original value
For example: (2,-3) <span>→ (6,-3)
Now since the Trapezoid was translated down 3,
Then all the (y)'s should be 3 less than the original value
For example: (2,-3) </span><span>→ (2,-6)
Now do this to all the Vertices
</span>(6,-6)(8,-6)(9,-5)(5,-5)
Final Answer: <span>Option 2</span>
Vinvika [58]3 years ago
7 0
<span>W′(6, −6), K′(8, −6), L′(9, −5) , and X′(5, −5)

</span>make me the brainiest answer 
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Write an equation for the line described in standard form with x-intercept 4 and y-intercept 5
stealth61 [152]

Answer:

Equation( \frac{5}{4}) x  + y = 5 is in the standard dorm.

Step-by-step explanation:

Here, the x - intercept = 4 , so ( 4,0) is a point on the line.

and the y - intercept = 5, so ( 0,5) is a point on the line.

So, the slope of the equation is given as  =  \frac{y_2 - y_1}{x_2-x_1 }  = \frac{5 -0}{0-4}  = -\frac{5}{4}

Now, the SLOPE INTERCEPT FORM of an equation is :

y  = mx + b: here m  = slope and b =  y- intercept

or, y =-( \frac{5}{4}) x + 5

Now, standard form is Ax +By = C

So, the standard form is ( \frac{5}{4}) x  + y = 5

Hence, the above equation ( \frac{5}{4}) x  + y = 5

is in the standard dorm.

4 0
2 years ago
Please please help me with this ​
iris [78.8K]

Answer:

angle FQE= Minor sector

6 0
2 years ago
Someone help me please
galben [10]

Answer:

∛27 = 3

Step-by-step explanation:

A radical is simply a fractional exponent: a^{(\frac{m}{n})} = \sqrt[n]{a^{m} }

Hence, ∛27 = 27^{(\frac{1}{3})}

Since 27 = 3³, then:

You could rewrite ∛27 as ∛(3)³.

\sqrt[3]{3^{(3)} } = 3^{[(3)*(\frac{1}{3})]}

Multiplying the fractional exponents (3 × 1/3) will result in 1 (because 3 is the <u><em>multiplicative inverse</em></u> of 1/3). The multiplicative inverse of a number is defined as a number which when multiplied by the original number gives the product as 1.

Therefore, ∛27 = 3.

3 0
2 years ago
Read 2 more answers
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