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3 years ago

5

Given the following diagram, find the missing measures. m o=m p m 1=36 m o=

1 answer:

Gala2k [10]3 years ago

4 0

If the diagram is this
-----------------------------------------------------------------------------------------------
https://media.glynlyon.com/g_geo_2013/3/page69a.gif
-----------------------------------------------------------------------------------------------
Then it would go like this

|dw:1392274742487:dw|

|dw:1392274778511:dw|

|dw:1392274826374:dw|

|dw:1392274862633:dw|

The answer would be 72

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Little late, but your answer would be 133 pieces of candy corn.

Best of luck.

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3 years ago

The table shows periodic data. Use the drop down menus to find the phase shift. y = 2 cosine(pi/4(x-c)) + 3. The function is shi

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3 years ago

Which of the following is true for the relation f(x) = 2x2 + 1?

guajiro [1.7K]

Neither the equation nor its inverse is a function

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3 years ago

Read 2 more answers

What are the solutions of the equation x4 + 3x2 + 2 = 0? Use u substitution to solve. x = plus-or-minus i StartRoot 2 EndRoot an

Rom4ik [11]

Answer:

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Step-by-step explanation:

We have been given an equation $x^4 + 3x^2 + 2 = 0$ . We are asked to find the solutions of our given equation using u-substitution.

We can rewrite our given equation as:

$(x^2)^2+3x^2+2 = 0$

Let us assume that $u=x^2$ .

$u^2+3u+2 = 0$

$u^2+2u+u+2 = 0$

$u(u+2)+1(u+2) = 0$

$(u+2)(u+1) = 0$

$(u+2)=0\text{ (or) }(u+1) = 0$

$u=-2\text{ (or) }u=-1$

Upon substituting back the value of u, we will get:

$x^2=-2\text{ (or) }x^2=-1$

$x=\pm\sqrt{-2}\text{ (or) }x=\pm\sqrt{-1}$

$x=\pm\sqrt{-1\cdot 2}\text{ (or) }x=\pm\sqrt{-1}$

Now we will use imaginary unit i. We know that $i^2=-1$ .

$x=\pm\sqrt{i^2\cdot 2}\text{ (or) }x=\pm\sqrt{i^2}$

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Therefore, the solutions for our given equation are $x=\pm\sqrt{2}i\text{ (or) }x=\pm i$ .

5 0

3 years ago

Read 2 more answers

Determine the quotient, q(x), and remainder, r(x) when

postnew [5]

ANSWER

Quotient:

$q(x) = 4 {x}^{2} + 4x + 4$

Remainder:

$r(x) = 2x + 6$

EXPLANATION

The given functions are

$f(x) = 12 {x}^{4} + 20 {x}^{3} + 40 {x}^{2} + 30x + 25$

and

$g(x) = 3 {x}^{2} + 2x + 5$

We want to determine the quotient and remainder when f(x) is divided by g(x).

We perform the long division as shown in the attachment.

The quotient is

$q(x) = 4 {x}^{2} + 4x + 4$

and the remainder is

$r(x) = 2x + 6$

6 0

3 years ago