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jarptica [38.1K]
2 years ago
5

What's the answer? Idk it bc I'm stupid and I don't feel like doing it

Mathematics
1 answer:
aliya0001 [1]2 years ago
4 0
H=-16x to the second power +136
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Which expression is equivalent to the following expression? 9(x - 5)
aleksley [76]

Answer:

9x - 45

Step-by-step explanation:

9 * x = 9x and 9 * -5 = -45. Therefore the answer is 9x - 45

7 0
2 years ago
Read 2 more answers
Patrick says that .500 is greater than .50 and .5 is he correct?
Dafna1 [17]
No he is not  .500, .50 and .5 are all the same number.

8 0
3 years ago
Work out m and c for the line:<br> 2x + 3y +1=0 <br><br>m=<br>c =​
Sergio [31]

Answer:

m = - \frac{2}{3} , c = - \frac{1}{3}

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

2x + 3y + 1 = 0 ( subtract 2x + 1 from both sides )

3y = - 2x - 1 ( divide terms by 3 )

y = - \frac{2}{3} x - \frac{1}{3} ← in slope- intercept form

with m = - \frac{2}{3} and c = - \frac{1}{3}

4 0
2 years ago
The figure below shows rectangle ABCD.
kipiarov [429]

Answer: We can find out the missing statement with help of below explanation.

Step-by-step explanation:

We have a rectangle ABCD with diagonals AC and BD ( shown in given figure.)

We have to prove: Diagonals AC and BD bisect each other.

In triangles, AED and BEC.

\angle ADB \cong \angle CBD ( By alternative angle theorem)

AD\cong BC ( Because ABCD is a rectangle)

\angle CAD\cong \angle ACB  ( By alternative angle theorem)

By ASA postulate,\triangle AED\cong \triangle BEC

By CPCTC, BE\cong ED and CE\cong EA

⇒ BE= ED and CE=EA

By the definition of bisector, AC and BD bisect each other.


4 0
3 years ago
Read 2 more answers
Which is the standard form of the equation of the parabola that has a vertex of (3, 1) and a directric of x = -2?
vagabundo [1.1K]
Check the picture below.  So,the parabola looks like so, notice the distance "p". since the parabola is opening to the right, then "p" is positive, thus is 5.

\bf \textit{parabola vertex form with focus point distance}\\\\&#10;\begin{array}{llll}&#10;\boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\&#10;(x-{{ h}})^2=4{{ p}}(y-{{ k}}) \\&#10;\end{array}&#10;\qquad &#10;\begin{array}{llll}&#10;vertex\ ({{ h}},{{ k}})\\\\&#10;{{ p}}=\textit{distance from vertex to }\\&#10;\qquad \textit{ focus or directrix}&#10;\end{array}\\\\

\bf -------------------------------\\\\&#10;\begin{cases}&#10;h=3\\&#10;k=1\\&#10;p=5&#10;\end{cases}\implies (y-1)^2=4(5)(x-3)\implies (y-1)^2=20(x-3)&#10;\\\\\\&#10;\cfrac{1}{20}(y-1)^2=x-3\implies \boxed{\cfrac{1}{20}(y-1)^2+3=x}

8 0
3 years ago
Read 2 more answers
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