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3 years ago

Corresponding angles form when a transversal intersects with two parallel lines.

Mathematics

1 answer:

Nadya [2.5K]3 years ago

3 0

Answer: True

When two lines are crossed by another line (which is called the Transversal), the angles in matching corners are called corresponding angles. When the two lines are parallel Corresponding Angles are equal.

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A bridge 48m tall.a scale model of the bridge 37cm long and 12cm tall .how long is the actual bridge .

Roman55 [17]

$\frac{48m}{x}$ = $\frac{37cm}{12cm}$

37x=48*12

37x=576

x=15.57m long.

The bridge is about 16 meters long.

6 0

3 years ago

Write an equation for the line passing through the given pair of points. give the final answer in standard form (-4,-2) and (3,-

Andrew [12]

Answer:

y = -2/7x - 22/7

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Slope-Intercept Form: y = mx + b

m - slope

b - y-intercept

Step-by-step explanation:

Step 1: Define

Point (-4, -2)

Point (3, -4)

Step 2: Find slope m

Substitute: $m=\frac{-4+2}{3+4}$
Add: $m=\frac{-2}{7}$

Step 3: Find y-intercept b

Define: y = -2/7x + b
Substitute: -4 = -2/7(3) + b
Multiply: -4 = -6/7 + b
Isolate b: -22/7 = b
Rewrite: b = -22/7

Step 4: Write linear equation

y = -2/7x - 22/7

7 0

2 years ago

Im too lazy to do this rn lol

gtnhenbr [62]

Answer:

You can drive at 53 miles per day.

Step-by-step explanation: Hope this helped )(-:

7 0

3 years ago

7.) The width of a vegetable garden is times its length. If the length of the garden is 73

Solnce55 [7]

Answer:

So, the length (L) = 7 3/4 ft = 31/4 (as an improper fraction)

the width (W) = (1/3)xL

So the width = 1/3 x (31/4) =

31/12 = 2 7/12 feet

Step-by-step explanation:

its up there

5 0

2 years ago

Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?

ivanzaharov [21]

Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.

8 0

3 years ago