Ask question

Ask question

All categories

3 years ago

15

Krystal and Danielle each improved their yards by planting daylilies and ivy. They bought their supplies from the same store. Kr

ystal spent $114 on 1 daylily and 12 pots of ivy. Danielle spent $150 on 10 daylilies and 10 pots of ivy. Find the cost of each item.

1 answer:

Eddi Din [679]3 years ago

6 0

Answer:

$9 and $6

Step-by-step explanation:

<u>Let the cost of daylilies = d and ivy = i, then we have below equations:</u>

d + 12i = 114
10d + 10i = 150

<u>Simplify the second equation:</u>

d + i = 15

<u>Subtract the second equation from the first:</u>

d - d + 12i - i = 114 - 15
11i = 99
i = 9, ivy costs $9

<u>Now find d:</u>

d = 15 - 9 = 6, daylilies cost $6

You might be interested in

Easy algebra question below first correct answer gets brainliest

Aliun [14]

Answer:

$x = \sqrt{5}$

Step-by-step explanation:

$3 {x}^{2} + 5 {x}^{2} = 35$

$8 {x}^{2} = 35$

${x}^{2} = 5$

$\sqrt{ {x}^{2} } = \sqrt{5}$

$x = \sqrt{5}$

8 0

3 years ago

The reyes family but for concert tickets for $252 what was the price per ticket

Bas_tet [7]

If you mean "The Reyes Family bought for concert tickets for $252",

Consider the total price is $252 and that there are four tickets. You are looking for the price of one ticket so view it as

Total Price = $252
Total Tickets = 4

Total Price divided by Total Tickets =
$\frac{252}{4}$
Then divide both sides by four because 4 divided by 4 is 1 ticket.
$\frac{252 \div 4}{4 \div 4} = \frac{63}{1}$

Then you should get 63 divided by 1.
63 divided by 1 is 63, So your answer is

"The price of one ticket that the Reyes Family bought is 63 dollars."

7 0

3 years ago

There is a single sequence of integers $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, $a_7$ such that \[\frac{5}{7} = \frac{a_2}{2!} + \frac

Nataliya [291]

You have a single sequence of integers $a_2,\ a_3,\ a_4,\ a_5,\ a_6,\ a_7$ such that

$\dfrac{a_2}{2!} + \dfrac{a_3}{3!} + \dfrac{a_4}{4!} + \dfrac{a_5}{5!} + \dfrac{a_6}{6!} + \dfrac{a_7}{7!}=\dfrac{5}{7},$

where $0 \le a_i < i$ for $i = 2, 3, \dots, 7.$

1. Multiply by 7! to get

$\dfrac{7!a_2}{2!} + \dfrac{7!a_3}{3!} + \dfrac{7!a_4}{4!} + \dfrac{7!a_5}{5!} + \dfrac{7!a_6}{6!} + \dfrac{7!a_7}{7!}=\dfrac{7!\cdot 5}{7},\\ \\7\cdot 6\cdot 5\cdot 4\cdot 3\cdoa a_2+7\cdot 6\cdot 5\cdot 4\cdot a_3+7\cdot 6\cdot 5\cdot a_4+7\cdot 6\cdot a_5+7\cdot a_6+a_7=6!\cdot 5,\\ \\7(6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)+a_7=3600.$

By Wilson's theorem,

$a_7+7\cdot (6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)\equiv 2(\mod 7)\Rightarrow a_7=2.$

2. Then write $a_7$ to the left and divide through by 7 to obtain

$6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6=\dfrac{3600-2}{7}=514.$

Repeat this procedure by $\mod 6:$

$a_6+6(5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5)\equiv 4(\mod 6)\Rightarrow a_6=4.$

And so on:

$5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5=\dfrac{514-4}{6}=85,\\ \\a_5+5(4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4)\equiv 0(\mod 5)\Rightarrow a_5=0,\\ \\4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4=\dfrac{85-0}{5}=17,\\ \\a_4+4(3\cdoa a_2+ a_3)\equiv 1(\mod 4)\Rightarrow a_4=1,\\ \\3\cdoa a_2+ a_3=\dfrac{17-1}{4}=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=\dfrac{4-1}{3}=1.$

Answer: $a_2=1,\ a_3=1,\ a_4=1,\ a_5=0,\ a_6=4,\ a_7=2.$

5 0

3 years ago

VO) --<br>I not know how to do this

masha68 [24]

Answer:

Square root and the power of 2 cancel each other. The answer remain 4/7

The square of a number is the number times itself. For example the square of 3 is 3x3. The square of 4 is 4x4.

The square root is just the opposite of the square. You can think of it as the "root" of the square or the number that was used to make the square.

8 0

3 years ago

1)calculate please it <br> 2) prove it please

Dvinal [7]

1. 2sin²(-2490) + tan(1410)
2(0.25) - 0.6
0.5 + 0.6
1.1

2. $\frac{2sin(\alpha)cos(\beta) - sin(\alpha - \beta)}{cos(\alpha - \beta) - 2sin(\alpha)sin(b)} = tan(\alpha + \beta)$
   $\frac{2sin(\alpha)cos(\beta) - sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)}{cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) - sin(\alpha)sin(\beta)} = tan(\alpha + \beta)$
   $\frac{sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)}{cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)} = tan(\alpha + \beta)$
   $\frac{sin(\alpha + \beta)}{cos(\alpha + \beta)} = tan(\alpha + \beta)$
   $tan(\alpha + \beta) = tan(\alpha + \beta)$

4 0

3 years ago