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steposvetlana [31]
3 years ago
11

Jane use 6 teaspoons and 10 cups of water if he wanted to maintain the same ratio how many teaspoons of salt would she need for

5 cups of water
Mathematics
1 answer:
balu736 [363]3 years ago
7 0

Answer: 3 teaspoons.

Step-by-step explanation: Notice that 5 cups is half of the original 10 cups, so the half of the original 6 teaspoons will be 3 teaspoons.

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Find the surface area and lateral surface area of the cylinder : radius: 3yd height 6yd
Svetradugi [14.3K]

Answer:

Step-by-step explanation:

lateral area=πr²h=π×3²×6=54π sq. yards

surface area=2πr²+πr²h=2π×3²+54π=18π+54π=72π sq. yards

4 0
3 years ago
At the park there a pool shaped like a circle. A - shaped path goes around the pool. Its inner diameter is 18 yd and its outer d
Hoochie [10]

Answer:

46 gallons

Explanation:

First, we need to calculate the area of the path. This area can be calculated as the difference between the area of the outer circle and the area of the inner circle. The area of a circle is equal to

A=\pi r^2

Where r is the radius. For the outer circle, the diameter is 28 yd, so the radius is equal to

radius = diameter/2

radius = 28 yd/2

radius = 14 yd

Then, the area of the outer circle is

\begin{gathered} A=(3.14)(14\text{ yd\rparen}^2 \\ \text{ A = \lparen3.14\rparen\lparen196 yd}^2) \\ A=615.44\text{ yd}^2 \end{gathered}

In the same way, the radius of the inner circle is

radius = diameter/2

radius = 18 yd/2

radius = 9 yd

Then, the area is

\begin{gathered} A=(3.14)(9\text{ yd\rparen}^2 \\ A=(3.14)(81\text{ yd}^2) \\ A=254.34\text{ yd}^2 \end{gathered}

So, the area of the path is

Area path = 615.44 yd² - 254.34 yd²

Area path = 361.1 yd²

Now, we know that one gallon can cover 8 yd², so we can calculate the number of gallons as

361.1\text{ yd}^2\times\frac{1\text{ gallon}}{8\text{ yd}^2}=45.14\text{ gallons}

Therefore, the number of gallons is 46 because we need to round the result to a whole number.

8 0
2 years ago
WILL GIVE BRAINLIST Examine the diagram and information to answer the question. A circle in the coordinate plane has a radius of
DanielleElmas [232]

Answer:  To find the equation of the circle, it is not necessary to find the legs of the triangle first.  

The length of the legs (equal) is 3√2 or  4.243

Step-by-step explanation:

The equation for a circle is similar to the Pythagorean Theorem

       a²   +   b²     = c²

( x - h )² + ( y - k )² = r², where ( h, k ) is the center and r is the radius.

The a and b terms are  changed to allow insertion of  h and k to locate the center of the circle on a Cartesian plane.

You can get the information needed from

Given: radius =6 and  "a center at the point (3,2)"

Substitute  3 for h and  2 for k  and  6 for r

( x - 3 )² + ( y - 2 )² = 6²    This is the equation of the circle .

From the diagram, it appears that x-3 = y-2

So, in the equation, we can substitute 2(x²) = 36

x² = 36/2 .   x² = 18

√x² = √18 = √9(2)

x = 3√2   =4.2426

the length of the horizontal leg is 4.243

Substitute for x in to find  x+3  the coordinate (x,2)

the coordinate x, 2 is  (7.24, 2)

Substitute for y in to find  y+2, the length of the vertical leg is 4.243.

y + 2 = 6.243  

So the coordinate (x,y) on the circle is (7.24, 6.24)

This does not account for Step 1 and  Step 2 in the question, matching expressions or equations to the steps. I hope you will be able to sort that our with the information here.

7 0
3 years ago
The answer to the problem
larisa86 [58]

She puts 8% of her salary in, so multiply her pay by 8% to get her yearly amount she saves:

45000 x 0.08 = $3,600 per year.

The company puts 6% of that in the account:

3600 x 0.06 = $216

So per year 3600 + 216 = $3,816 is saved.

3,816 x 2 = $7,632 is the 2 year total.

The answer is A.

6 0
4 years ago
Find the angle between the given vectors to the nearest tenth of a degree.
pochemuha
\bf \textit{angle between two vectors }\\ \quad \\
cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\\\
\theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\\\\
-----------------------------\\\\
\theta=cos^{-1}\left[ \cfrac{\ \textless \ 8,4\ \textgreater \ \quad \cdot \quad \ \textless \ 9,-9\ \textgreater \ }{(\sqrt{8^2+4^2})(\sqrt{9^2+(-9)^2})} \right]


\bf \theta=cos^{-1}\left[ \cfrac{(8\cdot 9)+(4\cdot -9)}{(\sqrt{64+16})(\sqrt{81+81})} \right]\implies \theta=cos^{-1}\left[ \cfrac{36}{(\sqrt{80})(\sqrt{162})} \right]
\\\\\\
\theta=cos^{-1}\left[ \cfrac{36}{(\sqrt{80})(\sqrt{162})} \right]\implies 
\theta=cos^{-1}\left[ \cfrac{36}{\sqrt{12960}}\right]
\\\\\\
\theta=cos^{-1}\left[ \cfrac{36}{36\sqrt{10}}\right]\implies \theta\approx 71.565^o
6 0
3 years ago
Read 2 more answers
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