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Marat540 [252]
1 year ago
8

Rita is planting saplings along her garden fence. When she started, she had x packages of saplings with 5 saplings per package.

She planted 15 saplings and has at most 20 left. How many packages of saplings did she start with? What would the solution look like on a number line graph?
When Rita started, she had
packages of saplings. The solution can be represented as
with an arrow to the
.
Mathematics
2 answers:
weqwewe [10]1 year ago
7 0

Answer:

If we represent x (number of packages of saplings) on a number line graph, it will start on number 4 (number of packages Rita still has) then moves to the right to number 7, that is the number of packages when Rita started to plant.

Step-by-step explanation:

Julli [10]1 year ago
6 0

Answer:

the answer is aproximatelay 20000

Step-byn:

adderal

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The 3rd degree Taylor polynomial for cos(x) centered at a = π 2 is given by, cos(x) = − (x − π/2) + 1/6 (x − π/2)3 + R3(x). Usin
Otrada [13]

Answer:

The cosine of 86º is approximately 0.06976.

Step-by-step explanation:

The third degree Taylor polynomial for the cosine function centered at a = \frac{\pi}{2} is:

\cos x \approx -\left(x-\frac{\pi}{2} \right)+\frac{1}{6}\cdot \left(x-\frac{\pi}{2} \right)^{3}

The value of 86º in radians is:

86^{\circ} = \frac{86^{\circ}}{180^{\circ}}\times \pi

86^{\circ} = \frac{43}{90}\pi\,rad

Then, the cosine of 86º is:

\cos 86^{\circ} \approx -\left(\frac{43}{90}\pi-\frac{\pi}{2}\right)+\frac{1}{6}\cdot \left(\frac{43}{90}\pi-\frac{\pi}{2}\right)^{3}

\cos 86^{\circ} \approx 0.06976

The cosine of 86º is approximately 0.06976.

8 0
2 years ago
N 2004, the General Social Survey (which uses a method similar to simple random sampling) asked, "Do you consider yourself athle
zmey [24]

<u>Answer-</u>

The standard error of the confidence interval is 0.63%

<u>Solution-</u>

Given,

n = 2373 (sample size)

x = 255 (number of people who bought)

The mean of the sample M will be,

M=\frac{x}{n} =\frac{255}{2373} =0.1075

Then the standard error SE will be,

SE=\sqrt{\frac{M\times (1-M)}{n}}

SE=\sqrt{\frac{0.1075\times (1-0.1075)}{2373}}=\sqrt{\frac{0.0959}{2373}}=0.0063=0.63\%

Therefore, the standard error of the confidence interval is 0.63%




6 0
2 years ago
silvia compro la misma licuadora que daniela, que costaba $355.5, pero como abonó con tarjeta de credito, le recargaron un 16%,
Debora [2.8K]

Answer:

Costo final= $412.38

Step-by-step explanation:

Dada la siguiente información:

Costo inicial= $355.5

Recargo de la tarjeta= 16% = 0.16

<u>Para calcular el costo final que debe pagar Silvia, debemos usar la siguiente información:</u>

Costo final= costo inicial*(1 + recargo)

Costo final= 355.5*1.16

Costo final= $412.38

4 0
1 year ago
Regular pentagonal tiles and triangular tiles are arranged in the pattern shown. The pentagonal tiles are all the same size and
professor190 [17]
First of all you have to find the missing measurements. The actual measurements for the angles in the hexagon are not given, but they give you an expression. You have to solve for x first so that you can plug it in and find the angle measurement. You have to equal the two sides that are given to you like this: 20x+48=33x+9. You solve for x and then plug it into each angle measurement. This should give you 108. Since it is a regular hexagon all of the sides are equal. If you look at the angle at the top of the hexagon you'll see two triangles and the angle. Since it lies on a straight line, it is all equal to 180. You already have the angle measurement of the hexagon and are missing the triangles. So 180-108=72. 72 is the missing part of the angle. You divide this by 2 in order to find each triangle angle measurements. the answer is 36 degrees. 
7 0
2 years ago
How much heat is required to raise the temperature of
Elina [12.6K]

Answer:

The amount of heat required to raise the temperature of liquid water is 9605 kilo joule .

Step-by-step explanation:

Given as :

The mass of liquid water = 50 g

The initial temperature = T_1 = 15°c

The final temperature = T_2  = 100°c

The latent heat of vaporization of water = 2260.0 J/g

Let The amount of heat required to raise temperature = Q Joule

Now, From method

Heat = mass × latent heat × change in temperature

Or, Q = m × s × ΔT

or, Q =  m × s × ( T_2 - T_1 )

So, Q = 50 g × 2260.0 J/g × ( 100°c - 15°c )

Or, Q =  50 g × 2260.0 J/g × 85°c

∴   Q = 9,605,000  joule

Or, Q =  9,605 × 10³ joule

Or, Q = 9605 kilo joule

Hence The amount of heat required to raise the temperature of liquid water is 9605 kilo joule . Answer

4 0
2 years ago
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