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2 years ago

The time of a race is 13.7seconds rounded to 1 decimal place.If the time was x seconds find the inequality in x

Mathematics

1 answer:

harkovskaia [24]2 years ago

7 0

Answer:

13.65 ≤ x < 13.75

Step-by-step explanation:

Let x represent the time in seconds. Since the time is 13.7 seconds when rounded to 1 decimal place, hence for this to be possible the minimum value of x is given as x ≥ 13.65 because 13.65 rounded to 1 decimal place is 13.7 seconds. Also, the maximum value of x is given as 13.75 because 13.75 rounded to 1 decimal place is 13.8 seconds.

Therefore the inequality to represent x is:

13.65 ≤ x < 13.75

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What is the constant of variation for the following table? If your answer is in the form of a fraction, write your answer in as

nikklg [1K]

Answer:

1/4

Step-by-step explanation:

Change is y (f9x) over the change in x.

Look at the f(x) column.

What is the change from -2 to -1? 1

What is the change from -1 to 0? 1

What is the change from 0 to 1? 1

What is the change from 1 to 2? 1

We see that this change is constant. It is always 1.

Look at the x column.

What is the change from -8 to -4? 4

What is the change from -4 to 0? 4

What is the change from 0 to 4? 4

What is the change from 4 to 8? 4

We see that the change is constant. It is always 1.

When we put this change the form a/b we get 1/4

8 0

1 year ago

Help I need to submit soon

Ostrovityanka [42]

Answer:

Step-by-step explanation:

I don't know if u want the explanation

8 0

3 years ago

Read 2 more answers

Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo

noname [10]

Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2} \right)}$

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it. Well, what is the slope of RS anyway?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}$

and let's check the reciprocal negative of that,

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6$

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}$

6 0

3 years ago

Find m∠TUL.

pogonyaev

Step-by-step explanation:

Actual graph for this problem is attached below

m∠TUV = 167°

m∠TUL = (x + 11)°

m∠LUV = (11x)°

m∠TUV= m∠TUL+ m∠LUV

now plug in the angles for each

m∠TUV= m∠TUL+ m∠LUV

$167= x+11 +11x$

solve the equation for x

$167= x+11 +11x\\167=11 +12x$

Subtract 11 from both sides

$156=12x$

divide both sides by 12

x=13

m∠TUL = (x + 11)°

m∠TUL = (13+ 11)° = (24)°

answer:

24°

4 0

3 years ago

Which equation can be solved using the expression StartFraction negative 3 plus-or-minus StartRoot (3) squared + 4 (10) (2) EndR

katen-ka-za [31]

Answer: B. 2 = 3x + 10x2

Step-by-step explanation:

This is the concept of quadratic equations; We required to find the type of equation that can be solved using the model that has been used to solve the equation such that the answer is:

[-3+-sqrt(3^2+4(10)(2))]/(2(10))

The formual that was applied here was a quadratic formula given by:

x=[-b+\-sqrt(b^2-4ac)]/2a

whereby from the our substituted values above,

a=10,b=3 and c=-2

such that the quadratic equation will be:

10x^2+3x-2

3 0

3 years ago

Read 2 more answers