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4 years ago

A transformation where a figure is turned around a fixed poing to create an image is called a

Mathematics

1 answer:

NemiM [27]4 years ago

8 0

A. reflection is defined as a figure and when moved causes a sudden change in a image

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Help i give 5 stars and a crown

REY [17]

Answer:

See below.

Step-by-step explanation:

1a)

Angles:

48, 90, 90, a.

Add:

48 + 90
= 138 + 90
= 228

Subtract:

360 - 228
= 132.

Angle a: 132.

1b)

To get angle h above 180°, We need to add h + 164:

h + 164
= 110 + 164
= 274.

h = 110.

This is how to get h over 180°.

6 0

2 years ago

75 percent of the sutdents in a school went on a field trip. If 285 students went on the trip, how any students are in the schoo

natita [175]

Answer:

380

Step-by-step explanation:

285 / x = 75 / 100

Cross multiply 28500 = 75x

Divide by 75 380

8 0

3 years ago

Can some one help me

svp [43]

Answer:

Domain: All real numbers, except for x=4

Range: All real numbers

Step-by-step explanation:

There is no limit to how far either of the graphs go, so it will be infinite.

4 0

3 years ago

Find the slope of the line

andrey2020 [161]

Answer:

-2/3x

Step-by-step explanation:

To find the slope, you would do rise/run.

To get to one point to the other, you move 2 units down and 3 to the right.

Hope this helps!

3 0

3 years ago

Read 2 more answers

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

3 years ago