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Svetach [21]
3 years ago
13

What is eighteen and three fourths divided by two and a half?​

Mathematics
1 answer:
MariettaO [177]3 years ago
4 0

Answer:

the answer is 7.5

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A. What is the slope of ABC?
givi [52]
A. 2/5
B.2/5
C. The slope of ABC and AGH are equal
5 0
3 years ago
Equation for ( 6, 13) ( 7900, 5 )
tia_tia [17]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

y=mx+b

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

(x_ {1}, y_ {1}) :( 6,13)\\(x_ {2}, y_ {2}): (7900,5)

We find the slope of the line:

m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {5-13} {7900-6} = \frac {-8} {7894} = - \frac {4} {3947}

Thus, the equation of the line is of the form:

y = - \frac {4} {3947} x + b

We substitute one of the points and find b:

13 = - \frac {4} {3947} (6) + b\\13 = - \frac {24} {3947} + b\\13+ \frac {24} {3947} = b\\b = \frac {51335} {3947}

Finally, the equation is:

y = - \frac {4} {3947} x + \frac {51335} {3947}

Answer:

y = - \frac {4} {3947} x + \frac {51335} {3947}

5 0
3 years ago
Can you help me? Please
Anni [7]

9514 1404 393

Answer:

  a.

Step-by-step explanation:

For an even-index radical, ...

  \sqrt[n]{x^n}=|x|\qquad\text{n even}

This lets us simplify the given expression as follows:

  \sqrt[4]{81x^6y^4}-|y|\sqrt[4]{x^6}-\sqrt[4]{16x^2}=\sqrt[4]{(3xy)^4x^2}-|y|\sqrt[4]{x^4x^2}-\sqrt[4]{2^4x^2}\\\\=3|xy|\sqrt[4]{x^2}-|xy|\sqrt[4]{x^2}-2\sqrt[4]{x^2}=\boxed{2|xy|\sqrt[4]{x^2}-2\sqrt[4]{x^2}}

4 0
2 years ago
What percent of 50 is 18?
JulijaS [17]

Answer:

36 %

Step-by-step explanation:

Convert the fraction to a decimal, then multiply by  100 .

3 0
3 years ago
Read 2 more answers
Which table describes the behavior of the graph of f(x) = 2x3 – 26x – 24?
In-s [12.5K]

Answer:

For x∈ {-∞,-3} y<0, below x-axis

x∈ {-3,-1} y>0, above x-axis

x∈ {-1,4} y<0, below x-axis

x∈ {4,∞} y>0, above x-axis

Step-by-step explanation:

f(x)=2x^{3}-26x-24

2x^{3}-26x-24=0

2x^{3}-26x-24=0\\2x^{3}-2x-24x-24=0\\2x(x^{2} -1)-24(x+1)=0\\2x(x+1)(x-1)-24(x+1)=0\\(x+1)(2x^{2} -2x-24)=0\\=> x+1=0 => x_{1}=-1\\

=> 2x^{2} -2x-24=0\\=>2(x^{2} -x-12)=0\\=> x^{2} -x-12=0\\=> x^{2} -4x+3x-12=0\\=> x(x-4)+3(x-4)=0\\=> (x-4)(x+3)=0\\

=> x-4=0\\=> x_{2}=4\\=> x+3=0\\=>x_{3}=-3\\

For x∈ {-∞,-3} y<0, below x-axis

x∈ {-3,-1} y>0, above x-axis

x∈ {-1,4} y<0, below x-axis

x∈ {4,∞} y>0, above x-axis



7 0
3 years ago
Read 2 more answers
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