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Dmitry_Shevchenko [17]
3 years ago
13

You buy 8 rose bushes and 15 daylilies for $193. Your friend buys 3 rose bushes and 15 daylilies for $117. Find the cost of each

daylily.​
Mathematics
1 answer:
RoseWind [281]3 years ago
4 0
2.35$ is the answer I think
You might be interested in
8x18÷4+15 use pemdas right answers only​
Gemiola [76]

Answer:

51

Step-by-step explanation:

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

Multiplication first, then division, and then addition.

(8*18)÷4+15 (Multiplication)

(144÷4)+15 (Division)

(36) + 15 (Addition)

   51

Hope that helps!

7 0
2 years ago
Read 2 more answers
Need an explanation
aleksandrvk [35]

Answer: Choice B

(6\sqrt{5}+5)i

==========================================================

Work Shown:

\sqrt{-5}+\sqrt{-25}+\sqrt{-125}\\\\\sqrt{-1*5}+\sqrt{-1*25}+\sqrt{-1*25*5}\\\\\sqrt{-1}*\sqrt{5}+\sqrt{-1}*\sqrt{25}+\sqrt{-1}*\sqrt{25}*\sqrt{5}\\\\i*\sqrt{5}+i*5+i*5*\sqrt{5}\\\\i*\sqrt{5}+5i+5i*\sqrt{5}\\\\(i\sqrt{5}+5i*\sqrt{5})+5i\\\\(\sqrt{5}+5\sqrt{5})i+5i\\\\(6\sqrt{5})i+5i\\\\(6\sqrt{5}+5)i\\\\

This points us to answer choice B

7 0
3 years ago
I need help what is 1/3x-3?​
lilavasa [31]

Answer:

\frac{1}{3}x-3=\frac{x}{3}-3

Step-by-step explanation:

\frac{1}{3}x-3\\\\\frac{1}{3}x = \frac{x}{3}  \\\\Find\:L.C.M.\\\\=\frac{x}{3}-3

8 0
3 years ago
An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i
Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = \sqrt{(15-x)^2+1^2}

The total time, 't', is given as follows;

t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}

The minimum value of 't' is found by differentiating with an online tool, as follows;

\dfrac{dt}{dx}  = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} =  \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }

At the maximum/minimum point, we have;

\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89

Therefore, the distance to run, x ≈ 14.96 mile

6 0
2 years ago
Which expression is greater than 1/2
polet [3.4K]
Anything greater than 1/2 can be shown as >1/2
8 0
3 years ago
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