Help !!!!!! what is the value of 100 C 98??

Evaluate.

xenn [34]

23.

3•(-3)^2-4

3•(-3)^2

Exponents first.

-3^2=9

3•9=27

27-4= 2:.

8 0

3 years ago

Please help!! Solve for n.

Komok [63]

Answer:

n = 9

Step-by-step explanation:

Create equivalent expressions in the equation that all have equal bases, then solve for n.

3 x 3 = 9

Therefore, n = 9

Hope this helps! :D

6 0

3 years ago

Read 2 more answers

PLZ HELP PLZ HELP The histogram below shows the homework grades from the last assignment in Mr. Smith's class.

KonstantinChe [14]

Answer:

Median, Interquartile range

Step-by-step explanation:

I took the test.

4 0

3 years ago

Draw out a two column proof for each problem below. Complete all problems on one page and upload ONE photo of the entire assignm

Hunter-Best [27]

Two or more triangles are congruent if on comparison, they have equal lengths of sides, and measure of angles.

Therefore, the required proofs for each question are shown below:

Problem 1:

Congruent triangles are triangles with equal lengths of corresponding sides and measures of internal angles.

Thus,

STATEMENT REASON

1. <NMQ ≅ <NPQ Any point on a perpendicular bisector

makes equal measure of angle with the

two ends of the line segment.

2. NQ ⊥ MP Definition of a line.

3. MQ ≅ PQ Equal segments of a bisected line.

4. MN ≅ PN Any point on a perpendicular bisector

is at the same distance to the

two ends of the line segment.

5. <MNQ ≅ <PNQ Equal measure of the bisected angle.

Problem 2:

A line segment is the shortest distance between two points.

STATEMENTS REASONS

1. m<PSR ≅ m<PSQ A perpendicular bisector is always at a right

angle to the bisected line segment.

2. m<RPS ≅ m<QPS Equal measure of the bisected angle.

3. RS ≅ QS Property of a bisected line segment.

4. PR ≅ PQ Any point on a perpendicular bisector

is at the same distance to the two ends of

the line segment.

For more clarifications on the perpendicular bisector of a line segment, visit: brainly.com/question/12475568

#SPJ1

3 0

2 years ago

1. Calculate the slope of the line between the two distinct points that create the hypotenuse for each of the three triangles sh

valentina_108 [34]

Calculate the equation for the line: y = mx + b

First, find the slope. Find two points, and use point-slope formula:

m = slope = (y₂ - y₁)/(x₂ - x₁)

Let (x₁ , y₁) = (0 , 0) & (x₂ , y₂) = (9, 6)

slope = (9 - 0)/(6 - 0) = 9/6

Plug in 9/6 (or a simplified version) for m

y = (3/2)x + b

Now, solve for b. Get a coordinate point to use in the equation. In this case, use (6, 4), in which x = 6, y = 4

4 = (3/2)(6) + b

Simplify.

4 = (3 x 6)/2 + b

4 = 3 x 3 + b

4 = 9 + b

Isolate the b. Note the equal sign, what you do to one side, you do to the other. Subtract 9 from both sides

4 (-9) = 9 (-9) + b

b = 4 - 9

b = -5

Plug in -5 for b

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y = (3/2)x - 5 is your equation (2.); 3/2 is your slope (1.)

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7 0

4 years ago