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2 years ago

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3. Susan invested $10,000 in an account that earned 3% interest, compounded monthly.

1 answer:

alisha [4.7K]2 years ago

3 0

A) total = 10,000 *(1+0.03/12)^(12*t)

B) t = 5 years

10,000 *(1+0.03/12)^(12*5 ) = $11,616.17

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Each side of a sandbox is 7 feet long. It will cost $1.00 per square foot to replace the sand in the sandbox. What would be the

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Answer: $7.00

Step-by-step explanation:

each foot is $1, so times 7 is 7 dollars.

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What is the length of XQY?

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Answer:

88

Step-by-step explanation:

First you must find the circumference of the entire circle and to do this you use 2πr. The radius in this case is PX=15. This means that the circumference is 30π. Now to find the arc length, we know that the measure of arc XY is 23 degrees, and since there are 360 total degrees in a circle, we must subtract 23 from 360 to get 337° as the measure of arc XQY. Now we have to find what portion of the circumference of the circle is in this arc, and to do this you divide the arc length (337°) by 360° and multiply by the circumference of the circle. Doing this would get you $\frac{337}{360} *30\pi$ which equals ~28. Finally, you have to multiply this by the value of pi and get approximately 88, which is your answer.

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3 years ago

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Someone please help me!!

kobusy [5.1K]

<h3>Answer:</h3>

A) Isosceles

E) Obtuse

<h3>Step-by-step explanation:</h3>

Ways to Define a Triangle

Triangles can be defined in two ways: by angles and by sides. Equilateral, isosceles, and scalene are based on side length. Acute, right, and obtuse are based on angle measurements. Triangle may only fall under one category for side length and one for angle measure (2 categories total).

Side Length

First, let's define equilateral, isosceles, and scalene.

Equilateral - All 3 sides of the triangle are congruent (equilateral are always acute angles).
Isosceles - 2 of the sides are congruent.
Scalene - There are no congruent sides; each side has a different length.

The triangle above has 2 congruent sides as shown by the tick marks on the left and right sides. This means the triangle is isosceles.

Angle Measurements

Now, let's define acute, right, and obtuse.

Acute - All 3 angles are less than 90 degrees; all angles are acute.
Right - 1 of the angles is exactly 90 degrees; it has a right angle.
Obtuse - 1 of the angles is greater than 90 degrees; there is an obtuse angle.

The largest angle in the triangle is 98 degrees, which is obtuse. This means that the triangle is obtuse.

7 0

1 year ago

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

---------------------------------------------------

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

---------------------------------------------------

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

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2 years ago