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Mandarinka [93]
3 years ago
9

Rhada has a 3-pound bag of clay. Her craft project requires 3 oz of clay for each batch of 4 ornaments.

Mathematics
1 answer:
maw [93]3 years ago
7 0

Answer:

64

Step-by-step explanation:

The first thing we should know is the equivalence between pound and ounce.

1 pound equals 16 ounces. Which means that Rhada has a total of:

3 * 16 = 48 ounces of clay.

We are told that for every 3 ounces of clay you can make 4 ornaments., therefore:

48 * 4/3 = 64 ornaments

Which means that you can with the amount of clay you have can make 64 ornaments.

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In 1985, the cost of clothing for a certain familiy was $620. In 1995, 10 years later, the cost of clothing for this famility wa
Sholpan [36]

Answer:

$848

Step-by-step explanation:

Calculation for the cost of this familiy's clothing in 1991

First step is to calculate the amount that is increase per year

Increase per year= ($1000-$620)/(1995-1985)

Increase per year= 380/10

Increase per year= $38

Now let y be :38*x + $620 and let x be 6 years (1991-1985)

Second step is to calculate the cost of the clothing in 1991

y = 38*6 + 620

y=228+620

y = $848

Therefore the cost of this familiy's clothing in 1991 will be $848

6 0
3 years ago
“Camilla makes and sells jewelry. She has 8,160 silver beads and 2,880 black beads to make necklaces. Each necklace will contain
Usimov [2.4K]
<h3>Answer:</h3>

96 necklaces

<h3>Step-by-step explanation:</h3>

To find out how many necklaces Camilla can make, we need to see how many times the required number of beads "goes into" the available number of beads.

Silver: 8160/85 = 96

Black: 2880/30 = 96

Camilla has enough beads to make 96 necklaces.

_____

<em>Comment on this answer</em>

Obviously, if one of the quotients is smaller than the other, Camilla can only make as many necklaces as are supported by the constraining resource.

Even if she had 3000 black beads (way more than 2880), she could still only make 96 necklaces (for example) because she would run out of silver beads making that 96th necklace. (There would be 120 black beads left over in that scenario.)

4 0
3 years ago
Which of the following is the equation for the graph shown?a. x^2/144+y^2/95=1b. x^2/144-y^2/95=1c. x^2/95+y^2/144=1d. x^2/95-y^
Andrews [41]

e follSOLUTION

Given the question in the image, the following are the solution steps to answer the question

STEP 1: Write the general equation of an ellipse

\frac{\mleft(x-h\mright)^2}{a^2}+\frac{(y-h)^2}{b^2^{}}=1

STEP 2: Identify the parameters

the length of the major axis is 2a

the length of the minor axis is 2b

\begin{gathered} 2a=24,a=\frac{24}{2}=12 \\ 2b=20,b=\frac{20}{2}=10 \end{gathered}

STEP 3: Get the equation of the ellipse

\begin{gathered} By\text{ substitution,} \\ \frac{(x-h)^2}{a^2}+\frac{(y-h)^2}{b^2}=1 \\ \frac{(x-0)^2}{12^2}+\frac{(y-0)^2}{10^2}=1=\frac{x^2}{144}+\frac{y^2}{100}=1 \end{gathered}

STEP 4: Pick the nearest equation from the options,

Hence, the equation of the ellipse in the image is given as:

\frac{x^2}{144}+\frac{y^2}{95}=1

OPTION A

8 0
1 year ago
A parabola can be drawn given a focus of
Volgvan

Answer:

y=-\frac{1}{4}(x-2)^{2}+9

Step-by-step explanation:

Any point on a given parabola is equidistant from focus and directrix.

Given:

Focus of the parabola is at (2,8).

Directrix of the parabola is y=10.

Let (x,y) be any point on the parabola. Then, from the definition of a parabola,

Distance of (x,y) from focus = Distance of (x,y) from directrix.

Therefore,

\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|

Squaring both sides, we get

(x-2)^{2}+(y-8)^{2}=(y-10)^{2}\\(x-2)^{2}=(y-10)^{2}-(y-8)^{2}\\(x-2)^{2}=(y-10+y-8)(y-10-(y-8))...............[\because a^{2}-b^{2}=(a+b)(a-b)]\\(x-2)^{2}=(2y-18)(y-10-y+8)\\(x-2)^{2}=2(y-9)(-2)\\(x-2)^{2}=-4(y-9)\\y-9=-\frac{1}{4}(x-2)^{2}\\y=-\frac{1}{4}(x-2)^{2}+9

Hence, the equation of the parabola is y=-\frac{1}{4}(x-2)^{2}+9.

4 0
3 years ago
Determine the first four terms of the sequence in which the nth term is...
Anon25 [30]

Answer:

1/5, 1/6, 1/7, 1/8

Step-by-step explanation:

The formula for the sequence is (n+3)!/ (n+4)!

The first terms uses n=1

a1 = (1+3)!/ (1+4)! = 4!/5! = (4*3*2*1)/(5*4*3*2*1) = 1/5


The first terms uses n=2

a2 = (2+3)!/ (2+4)! = 5!/6! = (5*4*3*2*1)/(6*5*4*3*2*1) = 1/6


The first terms uses n=3

a3 = (3+3)!/ (3+4)! = 6!/7! = (6*5*4*3*2*1)/(7*6*5*4*3*2*1) = 1/7


The first terms uses n=4

a4 = (4+3)!/ (4+4)! = 7!/8! = (7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1) = 1/8

7 0
3 years ago
Read 2 more answers
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