I can't figure this out......help!

A candy shop charges 8.99 a pound it's competitor charges 6.50 for 12 ounces which is a better deal and by how much

Gnom [1K]

Answer:

It is a better deal to do the 6.50 for 12 ounces

Step-by-step explanation:

The reason is because the is 16 ounces in a pound so you would do 8.99/16 and you get .56 per ounce but is you do 6.50/12 you get .54 per ounce which is a better deal.

Hope this helped!!

3 0

3 years ago

Prove the following identities:

siniylev [52]

The picture won’t load message me so i can answer !!

7 0

2 years ago

Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v

blagie [28]

U = ( -8 , -8)
v = (-1 , 2 )
the magnitude of vector projection of u onto v =
dot product of u and v over the magnitude of v = (u . v )/ ll v ll

ll v ll = √(-1² + 2²) = √5

u . v = ( -8 , -8) . ( -1 , 2) = -8*-1+2*-8 = -8
∴ (u . v )/ ll v ll = -8/√5
∴ the vector projection of u onto v = [(u . v )/ ll v ll] * [v/ ll v ll]

 = [-8/√5] * (-1,2)/√5 = ( 8/5 , -16/5 )

The other orthogonal component = u - ( 8/5 , -16/5 )
= (-8 , -8 ) - ( 8/5 , -16/5 ) = (-48/5 , -24/5 )

So, u as a sum of two orthogonal vectors will be

u = ( 8/5 , -16/5 ) + (-48/5 , -24/5 )

3 0

3 years ago

Read 2 more answers

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

3 years ago

Help? Thankyou so much

timurjin [86]

If Andrew were to rest, he would not be moving (walking). During such a period, the beginning distance would be the same as the ending distance.

In this graph there are two such time intervals: BC and DE. Notice how the graph is 'flat' on each; this 'flat' graph indicates a speed of zero.

6 0

3 years ago