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2 years ago

Cashews cost $6.75/lb. and Jeanna can spend at most $30 for them. How many pounds can she buy?

Mathematics

2 answers:

Vika [28.1K]2 years ago

6 0

Answer:

4.44 lb

Step-by-step explanation:

$6.75 ---> 1 lb

$1 ---> (1/6.75) lb

$30 ---> [$30* (1/6.75)] lb

$30 ---> 4.44

So she can buy most 4.44 lb of pounds

Step-by-step explanation:

pls branliest to me he/she have 210 me onley 4

Harman [31]2 years ago

4 0

Answer:

4.44 lb

Step-by-step explanation:

$6.75 ---> 1 lb

$1 ---> (1/6.75) lb

$30 ---> [$30* (1/6.75)] lb

$30 ---> 4.44

So she can buy most 4.44 lb of pounds

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Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value

vodomira [7]

Answer:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1, \frac{\sqrt{3}}{2} - 1$

Step-by-step explanation:

we are given two coincident points

$\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

since they are coincident points

$\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^{2} (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)$

By order pair we obtain:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}$

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

$\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}$

now substitute:

$\rm\displaystyle \displaystyle 4 \sin ^{2} (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2$

distribute:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

collect like terms:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^{2} ( \theta) = \sin( \theta) + 2$

rearrange:

$\rm\displaystyle \displaystyle 4 \sin ^{2}( θ) + 4 \cos ^{2} ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2$

by Pythagorean theorem we obtain:

$\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2$

cancel 4 from both sides:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2$

move right hand side expression to left hand side and change its sign:

$\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0$

factor out sin:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0$

factor out 2:

$\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0$

group:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0$

factor out -1:

$\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

divide both sides by -1:

$\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0$

by Zero product property we acquire:

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}$

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

$\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \neq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}$

divide both sides by 2:

$\rm\displaystyle \displaystyle \displaystyle \cos(θ) = \frac{1}{2}$

by unit circle we get:

$\rm\displaystyle \displaystyle \displaystyle θ= {60}^{ \circ} , {300}^{ \circ}$

so when θ is 60° a is:

$\rm \displaystyle a = \tan( {60}^{ \circ} ) - 2 - 3 \sin( {60}^{ \circ} ) + 2 \cos( {60}^{ \circ} )$

recall unit circle:

$\rm \displaystyle a = \sqrt{3} - 2 - \frac{ 3\sqrt{3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = - \frac{ \sqrt{3} }{2} - 1$

when θ is 300°

$\rm \displaystyle a = \tan( {300}^{ \circ} ) - 2 - 3 \sin( {300}^{ \circ} ) + 2 \cos( {300}^{ \circ} )$

remember unit circle:

$\rm \displaystyle a = - \sqrt{3} - 2 + \frac{3\sqrt{ 3} }{2} + 2 \cdot \frac{1}{2}$

simplify which yields:

$\rm \displaystyle a = \frac{ \sqrt{3} }{2} - 1$

and we are done!

disclaimer: also refer the attachment I did it first before answering the question

5 0

3 years ago

Kevin and Michelle traveled together by car and shared the driving. Kevin drove for 3.25 hours, then Michelle drove for 1.5 hour

jek_recluse [69]

Using the relation of velocity, distance and time, it is found that 35 represents Michelle's velocity.

-----------------------------

Velocity is distance divided by time, then:

$v = \frac{d}{t}$

Kevin drove for 3.25 hours, thus $t_1 = 3.25$ . The distance drove by Kevin is:

$d_1 = v_1t_1$

$d_1 = 3.25v_1$

Michelle drove for 1.5 hours, thus $t_2 = 1.5$ . The distance drove by Michelle is:

$d_2 = v_2t_2$

$d_2 = 1.5v_2$

The total distance is:

$d = d_1 + d_2$

$d = 3.25v_1 + 1.5v_2$

Comparing to the expression:

$3.25v_1 + 1.5v_2 = 3.25(42) + 1.5(35)$

We get that 42 is Kevin's speed and 35 is Michelle's speed.

A similar problem is given at brainly.com/question/24316569

8 0

2 years ago

Please give the measure of NPQ and explain how you got it

anygoal [31]

Answer:

Step-by-step explanation:

NQP is a triangle. It has 3 angles given.

We know

sum of 3 angles in a triangle is 180 degrees

We are Given:

∠N = 2x

∠Q = 2x + 2

∠P = 134

THus, we can write and solve for x first:

∠N + ∠Q + ∠P = 180

2x + 2x + 2 + 134 = 180

4x + 136 = 180

4x = 180 - 136

4x = 44

x = 44/4

x = 11

Now, measure of ∠NQP is 2x + 2, so substituting the value of "x", we have:

2x + 2 = 2(11) + 2 = 22 + 2 = 24 degrees

So,

∠NQP = 24°

Correct answer is C

4 0

3 years ago

4. Is the binomial that represents the length of the fish tank a factor of the polynomial that

Arlecino [84]

But I hope someone answers u 2920

5 0

3 years ago

Nth triangular number = 2 (n + 1)

mina [271]

Answer:

a) 2(n+1)

=2(12+1)

=2*13

=26

b) 2(n+1)

=2(50+1)

=2*51

=102

7 0

3 years ago