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CaHeK987 [17]
3 years ago
14

All you need is in the photo please help​

Mathematics
1 answer:
rosijanka [135]3 years ago
8 0

Answer:

50

Step-by-step explanation:

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Francesca is painting a picture of the purple flowers in her garden. She mixes paint to make the perfect color for each type of
aniked [119]

Find total ounces mixed for each flower:

Daisies = 2+5 = 7 ounces

Tulips = 3 +6 = 9 ounces

Now find the ratio of purple to white for each one:

Daisies = 5/7 = 0.714

Tulips = 6/9 = 0.666

Daisies have more purple so would be the darker ones.

8 0
3 years ago
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7/9+6/8+6/3=<br><br>I really need the answer<br>​
Neporo4naja [7]

Answer:

\frac{1}{30}

Step-by-step explanation:

Take it in steps. First, find 7/9+6. Then we'll find 8+6/3, and, finally, we'll divide the two answers.

1:

7/9+6 = 7/15

2:

8+6/3 = 14/3

3:

\frac{\frac{7}{15}}{\frac{14}{3}} or \frac{7/15}{14/3}

Then take that in chunks: 7/14 and 15/3.

7/14 = 1/2

15/3 = 5/1

Use those to rewrite it as \frac{1/5}{2/3}.

1/5 = .2

2/3 ≈ .6667 so we'll keep writing it as 2/3

\frac{\frac{.2}{2}}{3}

.2/2 = .1, so:

\frac{.1}{3} = \frac{1}{30}

3 0
3 years ago
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What is the product?
natta225 [31]
The answer is 4.644 * 10^9
8 0
4 years ago
The shaded region R in diagram below is enclosed by y-axis, y = x^2 - 1 and y = 3.
Lostsunrise [7]

Cross sections of the volume are washers or annuli with outer radii <em>x(y)</em> + 1, where

<em>y</em> = <em>x(y) </em>² - 1   ==>   <em>x(y)</em> = √(<em>y</em> + 1)

and inner radii 1. The distance between the outermost edge of each shell to the axis of revolution is then 1 + √(<em>y</em> + 1), and the distance between the innermost edge of <em>R</em> on the <em>y</em>-axis to the axis of revolution is 1.

For each value of <em>y</em> in the interval [-1, 3], the corresponding cross section has an area of

<em>π</em> (1 + √(<em>y</em> + 1))² - <em>π</em> (1)² = <em>π</em> (2√(<em>y</em> + 1) + <em>y</em> + 1)

Then the volume of the solid is the integral of this area over [-1, 3]:

\displaystyle\int_{-1}^3\pi y\,\mathrm dy = \frac{\pi y^2}2\bigg|_{-1}^3 = \boxed{4\pi}

\displaystyle\int_{-1}^3 \pi\left(2\sqrt{y+1}+y+1\right)\,\mathrm dy = \pi\left(\frac43(y+1)^{3/2}+\frac{y^2}2+y\right)\bigg|_{-1}^3 = \boxed{\frac{56\pi}3}

8 0
3 years ago
What is the solution for the inequality shown below W+4&gt;0
myrzilka [38]
The answer is: W>-4
I hope this helps
4 0
3 years ago
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