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3 years ago

Keep gettin this one wrong please help

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

5 0

Answer:

30 Nickels and 188 Pennies

Step-by-step explanation:

okay, so to set up the equation first, we have to assign each coin a variable, let's call them p and n:

P= number of pennies

N= number of nickels

the value of a penny is 1 cent, so 1P, and the value of a nickel is 5 cents, so 5N

The problem states that he has 218 coins, meaning that the total number of pennies and nickels adds up to 218:

P + N = 218

the total value of the coins is $3.38, so that would mean that 1P + 5N has to equal $3.38:

1P + 5N = 338

Okay, so now that we have our equations let's solve them using elimination:

we have to get a common coefficient between both equations, so let's multiply our first equation by 5:

P x 5 = 5P

N x 5 = 5N

218 x 5 = 1090

so, now we can solve by elimination:

5P + 5N = 1090

1P + 5N = 338

the N's cancel out:

4P = 752

divide both sides by 4:

P = 188

okay, so if theres a total of 218 coins, subtract 188 from 218:

218 - 188 = 30

so, there are 30 nickels and 188 pennies.

check our work:

5 x 30 = 150

1 × 188 = 188

150 + 188 = 338

338 = 338

I hope this helps! :)

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ozzi

Answer:

90 km, N 46° E

Step-by-step explanation:

A jet flies due North for a distance of 50 km and then on a bearing of N 70° E for a further 60 km. Find the distance and bearing of the jet from its starting point.

Look at the diagram I drew of this scenario. You can see the jet flies North for 50 km, and then turns at a 70° angle to fly another 60 km. We want to find the distance from the starting point, SP, to angle C (labeled).

This will be the jet's distance from its starting point.

In order to find the bearing of the jet from its starting point, we will need to find the angle formed between distances b and c, labeled angle A.

The Law of Cosines will allow us to use two known sides and one known angle to solve for the sides opposite of the known angle.

In this case, the known angle is 110° (angle B) so we will use the Law of Cosines respective to B.

b² = a² + c² - 2ac cosB

Substitute the known values into the equation and solve for b, the distance from the starting point (A) to the endpoint (C).

b² = (60)² + (50)² - 2(60)(50) cos(110°)
b² = 6100 -(-2052.12086)
b² = 8152.12086
b = 90.28909602
b ≈ 90 km

The distance of the jet from its starting point is 90 km. Now we can use this b value in order to calculate angle A, the bearing of the jet.

The Law of Cosines with respect to A:

a² = b² + c² - 2bc cosA

Substitute the known values into the equation and solve for A, the bearing from the starting point (clockwise of North).

(60)² = (90.28909602)² + (50)² - 2(90.28909602)(50) cosA
3600 = 8152.12086 - 6528.909602 cosA
-4552.12086 = -6528.909602 cosA
0.6972252853 = cosA
A = cos⁻¹(0.6972252853)
A = 45.79519
A ≈ 46°

The bearing of the jet from its starting point is N 46° E. This means that it is facing northeast at an angle of 46° clockwise from the North.

7 0

2 years ago

How to find the zeros of a non factorable function?

Mice21 [21]

Assuming it's for a polynomial function, this video explains a lot https://www.youtube.com/watch?v=OrLz7yide2g hope this helps:)

5 0

3 years ago

Can someone show me how to do this ? please :)) will award brainliest

sveticcg [70]

Answer:

x=16

Step-by-step explanation:

Because of Thales Intercept Theorem, AN/NG=NE/GL

NE/GL=1/2, AG=2x-9+x+7=3x-2

2x-9/3x-2=1/2

x=16

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3 years ago

How do i draw a decimal for example 0.26?

elixir [45]

First, you need convert the decimals into fraction

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The next step would be drawing 50 small boxes on a piece of paper. Make it colorless.

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3 years ago

Read 2 more answers

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$

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2 years ago