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3 years ago

What type of quadrilateral is DEFG? Select all that apply.

Mathematics

1 answer:

Ierofanga [76]3 years ago

5 0

Answer:

B. parallelogram

Step-by-step explanation:

Given

See attachment for quadrilateral

Required

Select the type of quadrilateral

From the attached image, we have:

$DE = FG$ -- Opposite, Parallel and Equal

$EF = GD$ -- Opposite, Parallel and Equal

Since all 4 sides are not equal, then it is neither a square; nor a rhombus

Hence, the shape is a parallelogram

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1. The beginning steps for determining the center and radius of a circle using the completing the square method are shown below:

svet-max [94.6K]

(1) Answer : $(x^2 - 6x + 9) + (y^2- 4y + 4) = 3 + (9 + 4)$

Step 1: $x^2 - 6x + y^2 - 4y = 3$

Step 2: $(x^2- 6x) + (y^2 - 4y) = 3$

In completing the square method we take coefficient of x and divide by 2 and the square it . Then add it on both sides

The coefficient of x is -6. $\frac{-6}{2}$ = (-3)^2 = 9

The coefficient of y is -4. $\frac{-4}{2}$ = (-2)^2 = 4

Step : $(x^2- 6x + 9) + (y^2 - 4y + 4) = 3 +9 + 4$

(2) $x^2 + y^2 + 6x - 6y + 2 = 0$

To find center and radius we write the equation in the form of

$(x-h)^2 + (y-k)^2 = r^2$ using completing the square form

Where (h,k) is the center and 'r' is the radius

$x^2 + y^2 + 6x - 6y + 2 = 0$

$(x^2 + 6x) + (y^2 - 6y) + 2 = 0$

In completing the square method we take coefficient of x and divide by 2 and the square it . Then add it on both sides

$(x^2 + 6x + 9) + (y^2 - 6y + 9) = -2 + 9 + 9$

$(x + 3)^2 + (x - 3)^2 = 16$

Here h= -3 and k=3 and $r^2 = 16$ so r= 4

Center is (-3,3) and radius = 4

Step 2: $(x^2 + 8x) + (y^2 - 6y) = 11$

Step 3: $(x^2 + 8x + 16) + (y^2 - 6y + 9) = 11 + (16 + 9)$

Step 4: $(x^2 + 8x + 16) + (y^2 - 6y + 9) = 36$

We factor out each quadratic

(x^2 + 8x + 16) = (x+4)(x+4) = $(x+4)^2$

((y^2 - 6y + 9)) = (x-3)(x-3) = $(x-3)^2$

Step 5 : $(x + 4)^2 + (y - 3)^2 = 6^2$

5 0

3 years ago

Let f(x) = x + 8 and g(x) = x2 − 6x − 7. Find f(g(2)). (1 point)

pantera1 [17]

Hello,

f(x)=x+8
g(x)=x²-6x-7
f(g(x))=f(x²-6x-7)=x²-6x-7+8=x²-6x+1

f(g(2))=2²-6*2+1=4-12+1=-8+1=-7

Answer A

4 0

3 years ago

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$

Thus the given are evaluated in terms of x and y

3 0

3 years ago

Find the measures of the missing angles (picture attached)

ella [17]

Here, sum of all the angles would be equal to 360.
We can see there are two right angles (=90)

So, 90 + 90 + x + 60 = 360
240 + x = 360
x = 360 - 240
x = 120

In short, Your Answer would be 120

Hope this helps!

4 0

3 years ago

What do you have to divide by to get from 60 to 50??

Rasek [7]

$60:50=\frac{60}{50}=\frac{6}{5}=\boxed{1\frac{1}{5}\ or\ decimal\ form\ 1\frac{1\cdot2}{5\cdot2}=1\frac{2}{10}=1.2}$

7 0

3 years ago

Read 2 more answers