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3 years ago

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HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP HELP I NEED HELP A

SAP HELP I NEED HELP ASAP HELP I NEED HELP ASAP

1 answer:

Andre45 [30]3 years ago

6 0

Answer:

D.) the amount in the account increase by 4% each year

Step-by-step explanation:

Compound interest formula: P*(1+r/n)^(nt)

P=initial principal balance

r=interest rate

n=times applied per [time (eg year)]

t=number of time periods gone by

The expression: 2500(1.04)^x

The expression: 2500(1+1/25)^(nt)

1.04=1+1/25=increases by 4% each year

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I’m on a timer PLZ HELP

MariettaO [177]

Answer:

answer is b

Step-by-step explanation:

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3 years ago

Which expressions are equivalent to 2^5•2^4? Check all that apply.

Maurinko [17]

Option A: $2^9$

Option E: $2^{-2}.2^{11}$

Option F: $(2.2.2.2.2)(2.2.2.2)$

Solution:

Given expression is $2^5.2^4$ .

To find which expression is equivalent to the given expression.

Option A: $2^9$

Using exponent rule: $a^b.a^c=a^{(b+c)}$

$2^5.2^4=2^{(5+4)}=2^9$

Therefore $2^9$ is equivalent to the given expression.

Option B: $2^{20}$

It is not equivalent to the given expression.

Option C: $2.2^9$

$2.2^9=2^{(1+9)}=2^{10}$

Therefore, It is not equivalent to the given expression.

Option D: $2^{10}.2^2$

$2^{10}.2^2=2^{(10+2)}=2^{12}$

Therefore, It is not equivalent to the given expression.

Option E: $2^{-2}.2^{11}$

$2^{-2}.2^{11}=2^{(2-11)}=2^9$

Therefore, It is equivalent to the given expression.

Option F: $(2.2.2.2.2)(2.2.2.2)$

$(2.2.2.2.2)(2.2.2.2)=2^5.2^4$

Therefore, It is equivalent to the given expression.

Hence option A, Option E and Option E are equivalent to the given expression.

3 0

4 years ago

Read 2 more answers

A child builds two wooden train sets. The path of one of the trains can be represented by the function y = 2cos2x, where y repre

Svet_ta [14]

Using a trigonometric equation, it is found that it will take 2.28 minutes until the two trains are first equidistant from the child.

<h3>What is the distance of each train to the child?</h3>

The distance of the first train is:

$y_1 = 2\cos^{2}x$

The distance of the second train is:

$y_2 = 3 + \cos{x}$

<h3>When are the trains equidistant to the child?</h3>

When $y_1 = y_2$ , hence:

$2\cos^{2}x = 3 + \cos{x}$

The following substitution is made:

$z = \cos{x}$

Hence:

$2z^2 = 3 + z$

$2z^2 - z - 3 = 0$

Which is a quadratic equation with coefficients $a = 2, b = -1, c = -3$ , hence:

$\Delta = b^2 - 4ac = (-1)^2 - 4(1)(-3) = 13$

$z_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{1 + \sqrt{13}}{4} = 1.5$

$z_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{1 - \sqrt{13}}{4} = -0.65$

Then, applying the trigonometric equation, considering that $-1 \leq z \leq 1$ due to the range of the cosine function:

$z_2 = \cos{x_2}$

$x_2 = \arccos{z_2} = \arccos{-0.65} = 2.28$

It will take 2.28 minutes until the two trains are first equidistant from the child.

You can learn more about trigonometric equations at brainly.com/question/2088730

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3 years ago

Mr. Richard's science class is conducting a variety of experiments with projectiles and falling objects. Two groups of students

liberstina [14]

The general equation for height is:
h = -0.5gt^2 + (v0)t + h0, where g = -9.8 m/s^2 or -32 ft/s^2, v0 = initial velocity, and h0 = initial height.
1) h = -4.9t^2 + 19: -4.9 means the units are in meters, and 19 means that the initial height was 19 m, which is group C.
2) -16t^2 means the units are in feet, and 50t means that the initial velocity was 50 ft/s upwards. This is group B.
3) -16t^2 again means the units are in feet, and 50 means the initial height was 50 ft. This is group D.
4) -4.9t^2 means the units are in meters, and 19 is the initial height of 19 m. This is for group A.

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3 years ago

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The triangles are similar.<br><br> What is the value of x?

qaws [65]

96/24 = 4

25*4=6x+28

72=6x

x=12

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3 years ago

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