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otez555 [7]
3 years ago
11

Jordan is chopping wood for the winter. For 10 days in a row, he follows the same process: In the morning, he chops 20 pieces of

wood, and in the evening, he chops an additional 30 pieces of wood. On the 11 day, Jordan chops only 5 pieces of wood. Write a numerical expression with parentheses for the total number of pieces of wood Jordan chopped. Do not simplify your expression.

Mathematics
2 answers:
Ronch [10]3 years ago
8 0

i would write this:

total = 10(20+30) +5

Bingel [31]3 years ago
6 0
10 x (20 + 30) + 5
10 x 50 + 5
500 + 5

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Which set of ordered pairs represents y as a function of x?
KatRina [158]

Answer:

(2,8), (3,5),

Step-by-step explanation:

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3 years ago
3n^5 divided by 6n^3 simplified,<br> thank
insens350 [35]

Answer:

\frac{ {n}^{2} }{2}

Step-by-step explanation:

\frac{3 {n}^{5} }{6 {n}^{3} }  =  \frac{3}{6}  \times  \frac{ {n}^{5} }{ {n}^{3} }  =  \frac{1}{2}  \times  {n}^{5 - 3}  = \frac{1}{2} \times {n}^{2} =  \frac{ {n}^{2} }{2}

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2 years ago
ASAP
pav-90 [236]

Answer:

+ 2 is a value for "C"

if the general form is ax^2 + bx +c and you want the A to be a positive integer...

then the c value would be -8

3x^2 - 8 = 0

Step-by-step explanation:

3 0
3 years ago
Object is thrown upward from a height of 15 ft at an initial vertical velocity of 30 ft per second. How long will it take to hit
dem82 [27]

Answer:

2.25 s.

Step-by-step explanation:

We'll begin by calculating the time taken for the object to get to the maximum height from the point of projection. This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₁) to reach the maximum height from the point of projection =?

v = u – gt₁ (since the object is going against gravity)

0 = 30 – (32.15 × t₁

0 = 30 – 32.15t₁

Collect like terms

0 – 30 = – 32.15t₁

– 30 = – 32.15t₁

Divide both side by – 32.15

t₁ = –30 / –32.15

t₁ = 0.93 s

Next, we shall determine the maximum height reached by the object from the point of projection.

This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Maximum height (h) reached from the point of projection =?

v² = u² – 2gh (since the object is going against gravity)

0² = 30² – (2 × 32.15 × h)

0 = 900 – 64.3h

Collect like terms

0 – 900 = – 64.3h

– 900 = – 64.3h

Divide both side by – 64.3

h = –900 / –64.3

h = 14 ft

Thus, the maximum reached by the object from the point of projection is 14 ft.

Next, we shall determine the height to which the of object is located from the maximum height reached to the ground. This can be obtained as follow:

Height (h₀) from which the object was projected = 14 ft

Maximum Height (h) reached from the point of projection = 14 ft

Height (hₗ) to which the of object is located from the maximum to the ground =?

hₗ = h₀ + h

hₗ = 14 + 14

hₗ = 28 ft

Thus, the height to which the of object is located from the maximum reached to the ground is 28 ft.

Next, we shall determine the time taken for the object to get to the ground from the maximum height reached. This can be obtained as follow:

Height (hₗ) to which the of object is located from the maximum to the ground = 28 ft

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₂) taken for the object to get to the ground from the maximum height reached =?

hₗ = ½gt₂²

28 = ½ × 32.15 × t₂²

28 = 16.075 × t₂²

Divide both side by 16.075

t₂² = 28 / 16.075

Take the square root of both side

t₂ = √(28 / 16.075)

t₂ = 1.32 s

Finally, we shall determine the time take for the object to get to the ground from the point of projection. This can be obtained as follow:

Time (t₁) to reach the maximum height from the point of projection = 0.93 s

Time (t₂) taken for the object to get to the ground from the maximum height reached = 1.32 s

Time (T) take for the object to get to the ground from the point of projection =?

T = t₁ + t₂

T = 0.93 + 1.32

T = 2.25 s.

Therefore, the time take for the object to get to the ground from the point of projection is 2.25 s.

7 0
2 years ago
Given the sample triangle below and the conditions , find the hypotenuse of the triangle
alukav5142 [94]
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</span>Make sure that your triangle is a right triangle.<span> The Pythagorean Theorem only works on right triangles, and by definition only right triangles can have a hypotenuse. If your triangle contains one angle that is exactly 90 degrees, it is a right triangle and you can proceed.</span><span>Right angles are often notated in textbooks and on tests with a small square in the corner of the angle. This special mark means "90 degrees."
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</span>Assign variables a, b, and c to the sides of your triangle.<span> The variable "c" will always be assigned to the hypotenuse, or longest side. Choose one of the other sides to be </span>a,<span> and call the other side </span>b<span> (it doesn't matter which is which; the math will turn out the same). Then copy the lengths of a and b into the formula, according to the following example:</span><span>If your triangle has sides of 3 and 4, and you have assigned letters to those sides such that a = 3 and b = 4, then you should write your equation out as: <span>32 + 42 = c2</span>.
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Find the squares of a and b.<span> To find the square of a number, you simply multiply the number by itself, so </span><span>a2 = a x a</span>. Find the squares of both a and b, and write them into your formula.<span><span>If a = 3, a2 = 3 x 3, or 9. If b = 4, then b2 = 4 x 4, or 16.</span><span>When you plug those values into your equation, it should now look like this: <span>9 + 16 = c2</span>.
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<span>Add together the values of <span>a2</span> and <span>b2</span>.</span><span> Enter this into your equation, and this will give you the value for c</span>2. There is only one step left to go, and you will have that hypotenuse solved!<span>In our example, 9 + 16 = 25, so you should write down <span>25 = c2</span>.

</span>
6 0
3 years ago
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