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MariettaO [177]
3 years ago
9

The average of the degree

Mathematics
1 answer:
maria [59]3 years ago
8 0
The answer is C.
Octagon is 135
And decagon is 144
So 144-135=11
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What is the scale factor of 24 and 18<br><br> 1/6<br> 2/3<br> 1/4<br> 3/4
Soloha48 [4]

Answer:

The answer is 3/4

Step-by-step explanation:

look up scale factor calculator

4 0
3 years ago
Write each equation in standard form using integers? <br><br> y = 3x + 1<br><br> y = 4x - 7
Maru [420]

Answer:

The standard form of y = 3 x + 1 is         -3x + y = 1 .

The standard form of y  = 4x - 7 is         -4x + y = -7.

Step-by-step explanation:

Here the given equation are:

y = 3 x + 1

and y  = 4x - 7

The above equations are of the form:  y = m x + C

Now, the STANDARD FORM of the equation is Ax + By  = C

<u>Consider equation 1: </u>

y = 3 x + 1

or, y - 3x  = 1

or. -3x + y = 1

⇒ The equation  -3x + y = 1  is in STANDARD FORM.

<u>Consider equation 2: </u>

y = 4 x - 7

or, y - 4x  = -7

or. -4x + y = -7

⇒ The equation -4x + y = -7 is in STANDARD FORM.

3 0
3 years ago
Given: p || q, and r || s.
MariettaO [177]

Answer:

A.

Statement: ∠6 ≅ ∠14

Reason: For parallel lines cut by a transversal, corresponding angles are congruent.

Step-by-step explanation:

In the figure attached, a plot of the problem is shown.

Given p || q and r is a transversal which cut p and q, then ∠1 ≅ ∠5 and ∠2 ≅ ∠6.

Given r || s and q is a transversal which cut r and s, then ∠6 ≅ ∠14 and ∠8 ≅ ∠16.

From the picture we see that ∠1 and ∠2 are supplementary, that is, their addition is equal to 180º. ∠2 ≅ ∠6 and ∠6 ≅ ∠14, then ∠2 ≅ ∠14, in consequence ∠1 and ∠14 are supplementary.

8 0
3 years ago
Read 2 more answers
I need help. Also need the answer in step by step form
Nutka1998 [239]

Answer:

x=2

Step-by-step explanation:

f(x)=-3x+1, f(x)=-5

-5=-3x+1, 3x=6, x=2. Answered by gauthmath

7 0
3 years ago
Read 2 more answers
I am equation of the line that passes through the point (2,3) with slope 3 please answer
Alinara [238K]

\huge{\boxed{y-3=m(x-2)}}

Point-slope form is y-y_1=m(x-x_1), where m is the slope and (x_1, y_1) is a point on the line.

Substitute the values. \boxed{y-3=m(x-2)}

<em>Note: This is in point-slope form. Let me know if you need a different form. Also, if you have any more problems similar to this, I encourage you to try them on your own, and ask on here if you are having trouble.</em>

7 0
3 years ago
Read 2 more answers
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