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2 years ago

14

Please help!!

1 answer:

dexar [7]2 years ago

8 0

Answer:C

Step-by-step explanation: You want to add up all the marbles, and then take the red marbles and divide by all the marbles 7/20

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Charlie gathers apples, pears, and plums. The numbers of apples and plums are 10-partners. There are the same number of apples a

aalyn [17]

Answer:

<u>The total number of pieces of fruit that Charlie can gather ranges from 11 to 19.</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Numbers of apples and plums are 10-partners, it means that the sum of apples and plums is 10.

Number of apples and pears are the same.

2. How many pieces of fruit can Charlie gather?

Apples = a

Plums = p

Pears = s

a + p = 10

Possible values of a and p:

1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5, 6 + 4, 7 + 3, 8 + 2, 9 + 1

Possible values of s:

1, 2, 3, 4, 5, 6, 7, 8, or 9

<u>Therefore, the total number of pieces of fruit that Charlie can gather ranges from 11 to 19.</u>

7 0

3 years ago

Find the value of x<br> 6x + 12<br> 3x - 3

Shalnov [3]

Answer:

19°

Step-by-step explanation:

<u>Concept</u><u> </u><u>Used </u><u>:</u><u>-</u>

The measure of a straight line is 180°. So the sum of angles on a straight line is 180° .

Using this ,

$\bf\implies 6x + 12 + 3x - 3 = 180\\\\\bf\implies 9x + 9 = 180\\\\\bf\implies 9x = 180-9 \\\\\bf\implies x =\dfrac{171}{9} \\\\\bf\implies\boxed{\red{\bf x = 19 }}$

<h3>Hence the value of x is 19°</h3>

3 0

3 years ago

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

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

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

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

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

8 0

2 years ago

Giving brainliest to first answer!!

almond37 [142]

Answer:

6

Step-by-step explanation:

6 0

3 years ago

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Find the length of the arc intercepted by central angle 0 = 250° in a circle with radius r =

andrew11 [14]

Answer:

6

Step-by-step explanation:

4 0

3 years ago