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Ilia_Sergeevich [38]
3 years ago
10

1)

Mathematics
1 answer:
AVprozaik [17]3 years ago
5 0
The answer to this is C.
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Help please due tomorrow
Nadusha1986 [10]

Answer:

Can you put a fraction. It is 1/3

7 0
3 years ago
Please I need Help!!!!!!!!!!!!!! with number 1 in scientific notation 5 x 10 to the forth power divided by 25 x 10 to the second
cluponka [151]

Answer:

200,000

Step-by-step explanation:

WE MUST USE THE PEMDAS RULE / ORDER OF OPERATION:

Parenthases

Exponent

Multiply

Divide

Add

Subtract

((5 x (10 to the 4th)) / 25) TIMES (10 to the second)

FIRST WE WORK ON THE FIRST PARENTHESIS:

10 to the 4th (10 x 10 x 10 x 10) = 10, 000

MULTIPLE 10, 000 by 5 (10,000 x 5 = 50, 000)

Then DIVIDE by 25 (50, 000 ÷ 25 is 2,000)

WE HAVE 2000 for parenthesis 1.

Now the second one:

10 to the 2nd in 10 x 10....thats equal to 100

2000 x 100 is 200,000! I HOPE YOU UNDERSTAND

~~~~~~~~~~~~~~

8 0
3 years ago
Read 2 more answers
I need a lot of help from the smartest people now right now
Dominik [7]

Answer:

K) I, II, and III

Step-by-step explanation:

Given the quadratic equation in standard form, <em>h </em>= -<em>at</em>² + <em>bt</em> + <em>c</em>, where <em>h </em>is the <u>height</u> or the projectile of a baseball that changes over time, <em>t</em>.  In the given quadratic equation, <em>c</em> represents the <u>constant term.</u> Altering the constant term, <em>c</em>, affects the <em>h</em>-intercept, the maximum value of <em>h</em><em>, </em>and the <em>t-</em>intercept of the quadratic equation.  

<h2>I. The <em>h</em>-intercept</h2>

The h-intercept is the value of the height<em>, h</em>, when <em>t = </em>0. This means that setting <em>t</em> = 0 will leave you with the value of the constant term. In other words:

Set <em>t</em> = 0:

<em>h </em>= -<em>at</em>² + <em>bt</em> + <em>c</em>

<em>h </em>= -<em>a</em>(0)² + <em>b</em>(0) + <em>c</em>

<em>h</em> = -a(0) + 0 + <em>c</em>

<em>h</em> = 0 + <em>c</em>

<em>h = c</em>

Therefore, the value of the h-intercept is the value of c.

Hence, altering the value of<em> c </em>will also change the value of the h-intercept.

<h2>II. The maximum value of <em>h</em></h2>

The <u>maximum value</u> of <em>h</em> occurs at the <u>vertex</u>, (<em>t, h </em>). Changing the value of <em>c</em> affects the equation, especially the maximum value of <em>h. </em>To find the value of the <em>t</em>-coordinate of the vertex, use the following formula:

<em>t</em> = -b/2a

The value of the t-coordinate will then be substituted into the equation to find its corresponding <em>h-</em>coordinate. Thus, changing the value of <em>c</em> affects  the corresponding <em>h</em>-coordinate of the vertex because you'll have to add the constant term into the rest of the terms within the equation. Therefore, altering the value of <em>c</em> affects the maximum value of <em>h.</em><em> </em>

<h2>III. The <em>t-</em>intercept</h2>

The <u><em>t-</em></u><u>intercept</u> is the point on the graph where it crosses the t-axis, and is also the value of <em>t</em> when <em>h</em> = 0. The t-intercept is the zero or the solution to the given equation. To find the <em>t</em>-intercept, set <em>h</em> = 0, and solve for the value of <em>t</em>.  Solving for the value of <em>t</em> includes the addition of the constant term, <em>c</em>, with the rest of the terms in the equation.  Therefore, altering the value of <em>c</em> also affects the<em> </em><em>t-intercept</em>.

Therefore, the correct answer is <u>Option K</u>: I, II, and III.

5 0
2 years ago
1/3y+1/4=5/12<br> can you tell me a quick way to solve this problem. thank you
Readme [11.4K]
<span> Assume that the expression is..

y/3 + ¼ = 5/12

Multiply both sides by 12 to get..

4y + 3 = 5

Then...

4y = 5 - 3
4y = 2
y = 2/4
y = ½

I hope this helps! </span>
6 0
3 years ago
I NEED HELP ON THESE QUESTIONS QUICKLY PLEASE!!!!
Korvikt [17]

The objective function Z = 2x + 3y cannot be maximized, while the maximum value of the objective function Z = 2x + 5y is 19

<h3>Part A (Question 1): Maximize Z</h3>

The objective function is given as:

Z = 2x + 3y

The constraints are given as:

3x+2y≤15

4x+y≤0

x≥0; y≥0

See attachment for the graph of the constraints

From the attached graph, we have the optimal solution to be:

(x,y) = (-3,12)

Because x≥0 & y≥0, and x is negative in (-3,12), then

The function cannot be maximized.

<h3>Part B (Question 7): Maximize Z</h3>

The objective function is given as:

Z = 2x + 5y

The constraints are given as:

x + y ≤ 5

y - x ≤ 1

x≥0; y≥0

See attachment for the graph of the constraints

From the attached graph, we have the optimal solution to be:

(x,y) = (2,3)

Substitute 2 and 3 for x and y in the objective function.

Z = 2 * 2 + 5 * 3

Evaluate

Z = 19

Hence, the maximum value of the objective function is 19

The questions 21, 22, 23 and 25 are not included in the attachment

Read more about maximized functions at:

brainly.com/question/16826001

#SPJ1

8 0
2 years ago
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