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3 years ago

Which construction is illustrated above?

Mathematics

2 answers:

Leni [432]3 years ago

6 0

Answer:

The construction is a parallel line to a given line from a point not on the line.

Step-by-step explanation:

For a given figure,

To construct a parallel line

Step 1 : Draw transversal line through any point not on line which is given.

Step 2: Using construction copy an angle, construct same angle formed by transversal line and given line.

Step 3: This is the last step in which when the copy of an angle completed draw a line parallel to the line.

grandymaker [24]3 years ago

5 0

Pretty sure it is A - a parallel line to a given line from a point not on the line

I'm working on the same test :)

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Dylan and his friends are going into a bakery. How much would it cost to buy 1 cookie and 5 brownies if each cookie costs $0.50

mixer [17]

Answer:

a)$ 11.75

b) $ 129.25

Step-by-step explanation:

Dylan and his friends are going into a bakery.

Each cookie costs $0.50 and each brownie costs $2.25?

a) How much would it cost to buy 1 cookie and 5 brownies

1 × $0.50 + 5 × $2.25

= $0.50 + 11.25

= $ 11.75

b) How much would it cost to buy 11 cookies and 55 brownies

11 × $0.50 + 55 × $2.25

= $5.5 + $123.75

= $ 129.25

4 0

3 years ago

46 divided by 2990 show me in long division form

zheka24 [161]

Answer:

hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

Step-by-step explanation:

8 0

3 years ago

(3x^2+4x-2)-(4x^2-2x-5)

stiv31 [10]

-x^2+6x+3
hope this helps :)

4 0

3 years ago

Read 2 more answers

Write out each step of the 3-step test for continuity for the following functions at the given point. If the function is discont

balandron [24]

If we simply substitute the value x=2 in the expression we have

$f(2) = \dfrac{4-12+8}{4-2-2} = \dfrac{0}{0}$

which is undefined.

But if we factor both numerator and denominator, we have

$f(x)=\dfrac{x^2-6x+8}{x^2-x-2} = \dfrac{(x-2)(x-4)}{(x+1)(x-2)}$

Since we are studying the limit as x approaches 2, we can assume that x is not 2. In this case, we can simplify the (x-2) parenthesis, and the expression becomes

$f(x)=\dfrac{x-4}{x+1}$

And we can evaluate this at 2 with no problems:

$f(2) = \dfrac{2-4}{2+1} = -\dfrac{2}{3}$

So, we have

$\displaystyle \lim_{x\to 2}f(x) = -\dfrac{2}{3}$

This means that in this case both left and right limits exist and are the same, so the limit exists, but the function is not defined at x=2. This is a removable discontinuity, because we can define the function as its limit, and we have a continuous function at x=2:

$f(x) = \begin{cases}\dfrac{x^2-6x+8}{x^2-x-2} &\text{if }x \neq 2\\ -\frac{2}{3} &\text{if }x = 2\end{cases}$

8 0

4 years ago

Please help me. i don’t want to fail

rosijanka [135]

Answer: i’m pretty sure the answer is wx, xy, yw

Step-by-step explanation:

3 0

3 years ago