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forsale [732]
3 years ago
10

Marcy needs to make 5 lb of a cashew/raisin mixture for a party. Her local grocery store sells raisins for $4.50/lb and cashews

for $9.50/lb. How many pounds of raisins and cashews should she buy if she wants to spend $40? Enter your answers in the boxes. raisins: lb cashews: lb
Mathematics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

Pounds of raisins = 1.5 lb

Pounds of cashew = 3.5 lb

Step-by-step explanation:

Let

Pounds of raisins = x

Pounds of cashew = y

x + y = 5 (1)

4.50x + 9.50y = 40 (2)

From (1)

x = 5 - y

Substitute x = 5 - y into

4.50x + 9.50y = 40

4.50(5 - y) + 9.50y = 40

22.50 - 4.50y + 9.50y = 40

- 4.50y + 9.50y = 40 - 22.50

5y = 17.5

y = 17.5 / 5

= 3.5

y = 3.5lb

Substitute y = 3.5 into

x + y = 5

x + 3.5 = 5

x = 5 - 3.5

= 1.5

x = 1.5lb

Pounds of raisins = 1.5 lb

Pounds of cashew = 3.5 lb

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Answer:

y = 70 and x = 20

Step-by-step explanation:

180 - 40 = 140

140/2 = 70

70 + 90 = 160

180 - 160 = 20

3 0
3 years ago
Polygon ABCD is rotated 90 degrees counterclockwise about the origin to create polygon
Aleonysh [2.5K]

answer with an example is attached below

4 0
3 years ago
The junior and senior students at Mathville High School are going to present an exciting musical entitled, "Math, What is it Goo
xeze [42]

Some parts are missing in the queston. Find attached the picture with the complete question

Answer:

                   \large\boxed{\large\boxed{161}}

Explanation:

Let's put the information in a table step-by step.

                                                (number of remaining students)

                                                        Juniors          Seniors

Condition

  • Initially                                           J                     S
  • 15 seniors left                                                   S - 15
  • Twice juniors as seniors         2(S - 15)
  • 3/4 of the juniors left              1/4×2(S - 15)
  • 1/3 of seniors left                                             2/3×(S - 15)

At the end, there were 8 more seniors than juniors:

  • 2/3×(S - 15) -  1/4×2(S - 15) = 8

Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

2/3\times (S - 15) -  1/4\times 2(S - 15) = 8

  • Mutilply all by 12:

8(S - 15)-6(S - 15)=96

  • Distribution property:

8S-120-6S-90=96

  • Addtion property of equalities:

8S-6S=96+120+90

  • Add like terms:

2S=306

  • Division property of equalities:

S=306/2=153

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

2/3\times (S-15)+1/4\times 2(S-15)

Just substitute S with 153 fo find the number of students that remained to perfom in the musical:

          2/3\times (153-15)+1/4\times 2(153-15)\\ \\ 2/3(138)+1/2(138)

          161

5 0
3 years ago
a company is offering a job with a salary of $30,000 for the first year and a 5% raise each year after that. if the 5% raise con
Ber [7]
<span>First year is 30,000.
</span><span>Earn 5% raise every year.
</span>Growth factor is 1.05 this sounds a lot like a geometric ratio.

An = A1 × r^(n-1)Sn = A1 × (1 - r^n) / (1-r)<span>n = 40
A1 = 30,000
A40 = $30,000 * 1.05^39 = $201,142.53

S40 = A1 * (1 - 1.05^40) / (1-1.05) which becomes:
S40 = 30,000 * -6.039988712 / -.05 which becomes: 
</span>S40 = -181,199.6614 / -.05 which becomes:
S40 = $3,623,993.227

<span>The individual yearly calculations are shown below: </span>

8 0
4 years ago
X = ? y = ? 16 45 degrees
mezya [45]

Let's put more details in the figure to better understand the problem:

Let's first recall the three main trigonometric functions:

\text{ Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}\text{ Tangent }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Adjacent Side}}

For x, we will be using the Cosine Function:

\text{ Cosine }\theta\text{ = }\frac{\text{ Adjacent Side}}{\text{ Hypotenuse}}Cosine(45^{\circ})\text{ = }\frac{\text{ x}}{\text{ 1}6}(16)Cosine(45^{\circ})\text{ =  x}(16)(\frac{1}{\sqrt[]{2}})\text{ = x}\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}\text{ 8}\sqrt[]{2}\text{ = x}

Therefore, x = 8√2.

For y, we will be using the Sine Function.

\text{  Sine }\theta\text{ = }\frac{\text{ Opposite Side}}{\text{ Hypotenuse}}\text{ Sine }(45^{\circ})\text{ = }\frac{\text{ y}}{\text{ 1}6}\text{ (16)Sine }(45^{\circ})\text{ =  y}\text{ (16)(}\frac{1}{\sqrt[]{2}})\text{ = y}\text{ }\frac{16}{\sqrt[]{2}}\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{2}}\text{ = }\frac{16\sqrt[]{2}}{2}\text{ 8}\sqrt[]{2}\text{ = y}

Therefore, y = 8√2.

5 0
1 year ago
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