All categories

2 years ago

Please help fast!!! Part A How many solutions does the pair of equations for lines A and B have?

Mathematics

1 answer:

amm18122 years ago

7 0

Answer:

Part A: 1

Explain: The solution for a pair of lines is where they intersect.

Part B: (3,4)

Explain: (3,4) is the place where the lines intersect.

You might be interested in

Solve this problem using the proper formula. A carpenter is building a bookcase the perimeter of the top is 108 in. What length

lapo4ka [179]

Assuming the length is y and the width is x
Perimeter = y + y + x + x
                = 2y +2x
                = 2 (y + x)
if the is 8 and the perimeter is 108
   108 = 2 (y +8)
    108/2 = y+8
    54    = y +8
    54 - 8 = y
    46 = y

Length = 46 inches

3 0

3 years ago

Read 2 more answers

Help! See the question below!

n200080 [17]

It’s the 3rd problem:)

5 0

3 years ago

Read 2 more answers

SOMEONE PLEASE HELP ASAP

Fittoniya [83]

Yes it proportional because they all function of y=3x
0x3=0
2x3=6
4x3=12
6x3=18

6 0

3 years ago

AN ICE CREAM STORE IS CONDUCTING A SURVEY ASKING CUSTOMERS TO RANK THE TOP 5 OUT OF 10 NEW FLAVORS INTRODUCED THIS YEAR. HOW MAN

katrin2010 [14]

The problem given above can be solved through the concept of permutation because of the importance of the arrangements. This may be solved through calculators directly by keying in,
10P5
That is, the permutation of 10 taken 5. The answer to this permutation is 30,240.

3 0

2 years ago

This is a geometry question, i need something quickly :)

Marysya12 [62]

Answer:

hope it helps mark me brainlieast!

Step-by-step explanation:

For triangle ABC with sides a,b,c labeled in the usual way,

c2=a2+b2−2abcosC

We can easily solve for angle C .

2abcosC=a2+b2−c2

cosC=a2+b2−c22ab

C=arccosa2+b2−c22ab

That’s the formula for getting the angle of a triangle from its sides.

The Law of Cosines has no exceptions and ambiguities, unlike many other trig formulas. Each possible value for a cosine maps uniquely to a triangle angle, and vice versa, a true bijection between cosines and triangle angles. Increasing cosines corresponds to smaller angles.

−1≤cosC≤1

0∘≤C≤180∘

We needed to include the degenerate triangle angles, 0∘ and 180∘, among the triangle angles to capture the full range of the cosine. Degenerate triangles aren’t triangles, but they do correspond to a valid configuration of three points, namely three collinear points.

The Law of Cosines, together with sin2θ+cos2θ=1 , is all we need to derive most of trigonometry. C=90∘ gives the Pythagorean Theorem; C=0 and C=180∘ give the foundational but often unnamed Segment Addition Theorem, and the Law of Sines is in there as well, which I’ll leave for you to find, just a few steps from cosC= … above. (Hint: the Law of Cosines applies to all three angles in a triangle.)

The Triangle Angle Sum Theorem, A+B+C=180∘ , is a bit hard to tease out. Substituting the Law of Sines into the Law of Cosines we get the very cool

2sinAsinBcosC=sin2A+sin2B−sin2C

Showing that’s the same as A+B+C=180∘ is a challenge I’ll leave for you.

In Rational Trigonometry instead of angle we use spreads, squared sines, and the squared form of the formula we just found is the Triple Spread Formula,

4sin2Asin2B(1−sin2C)=(sin2A+sin2B−sin2C)2

true precisely when ±A±B±C=180∘k , integer k, for some k and combination of signs.

This is written in RT in an inverted notation, for triangle abc with vertices little a,b,c which we conflate with spreads a,b,c,

(a+b−c)2=4ab(1−c)

Very tidy. It’s an often challenging third degree equation to find the spreads corresponding to angles that add to 180∘ or zero, but it’s a whole lot cleaner than the trip through the transcendental tunnel and back, which almost inevitably forces approximation.

6 0

2 years ago