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3 years ago

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A 2-column table with 5 rows. The first column, number of snapdragons, x, has the entries 11, 12, 13, 14. The second column, num

ber of daisies, y, has the entries 34, 33, 32, 31.
Hans is planting a garden with snapdragons and daisies. The table shows some possible combinations of the two plants. If Hans plants 29 daisies, how many snapdragons will he plant?

2 answers:

mihalych1998 [28]3 years ago

7 0

Answer:

x=45-y and 16

Step-by-step explanation:

mamaluj [8]3 years ago

3 0

9514 1404 393

Answer:

16

Step-by-step explanation:

The values in each row of the table have a total of 45. If Hans plants 29 daisies, he likely will plant ...

45 -29 = 16 . . . snapdragons

__

One can also find the answer by extending the table 2 more rows. Each x entry increases by 1 as each y entry decreases by 1. Then the number of snapdragons listed in the second added row will be 14+2 = 16.

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3 years ago

Angle $eab$ is a right angle, and $be = 9$ units. what is the number of square units in the sum of the areas of the two squares

Alla [95]

Let the measure of side AB be x, then, the measue of side AE is given by

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Also, AEFG is a square of size $\sqrt{9^2-x^2}$ , thus, the area of square AEFG is given by

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8 0

3 years ago

A = 6 and b = 24 work out the values a) a+b b) ab c) b/a d) d) (a+b)^2

Oksanka [162]

Answer:

a) 30

b) 144

c) 4

d) 900

Step-by-step explanation:

a) 6 + 24 = 30

b) 6 * 24 = 144

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5 0

3 years ago

Read 2 more answers

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:

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

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3 years ago

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oee [108]

It means to break down the number 20

6 0

3 years ago

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