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3 years ago

Please help with math asap: Rewrite without parentheses and simplify (y+7)^2

Mathematics

1 answer:

Andrew [12]3 years ago

4 0

y * y = y^2
y * 7 = 7y
7 * y = 7y
7 * 7 = 49

So we have:

y^2 + 7y + 7y + 49

Combine like terms:

y^2 + 14y + 49

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Suppose that y varies directly with x and y=5 when x=6. What is y when x=5? Be sure to simplify your answer.

Charra [1.4K]

Answer:

y = $\frac{25}{6}$

Step-by-step explanation:

Given y varies directly with x then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition y = 5 when x = 6

k = $\frac{y}{x}$ = $\frac{5}{6}$

y = $\frac{5}{6}$ x ← equation of variation

When x = 5, then

y = $\frac{5}{6}$ × 5 = $\frac{25}{6}$

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3 years ago

Please help meeeeeee. image attached

Shtirlitz [24]

The midpoint between -4 and 8 is answer B) 2

8 0

3 years ago

What is the equation of a parabola with (−2, 4) as its focus and y = 6 as its directrix? Enter the equation in the box.

SOVA2 [1]

Notice the picture below

the directrix is above the focus point, meaning the parabola is vertical and is opening downwards

now, "p" is the distance from the vertex to the focus point or the directrix, so that means, the vertex is between those two fellows, over the axis of symmetry, x = -2, since "p" is 1 unit, that puts the vertex at -2,5

since the parabola is opening downwards, that means the "p" value is negative, so is -1

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

$\bf \begin{cases}
p=-1\\
h=-2\\
k=5
\end{cases}\implies (x-(-2))^2=4(-1)(y-5)
\\\\\\
(x+2)^2=-4(y-5)\implies 
-\cfrac{1}{4}(x+2)^2=y-5
\\\\\\
\boxed{-\cfrac{1}{4}(x+2)^2+5=y}$

4 0

3 years ago

I need help ASAP! It's urgent.. PLISSSSS

natali 33 [55]

Answer:

a) 6 mins

b) 70km/h

c) t= 45

Step-by-step explanation:

a) The bus stops from t=10 to t=16 minutes since the distance the busvtravelled remained constant at 15km

Duration

= 16 -10

= 6 minutes

b) Average speed

= total distance ÷ total time

Total time

= 24min

= (24÷60) hr

= 0.4 h

Average speed

= 28 ÷0.4

= 70 km/h

c) Average speed= total distance/ total time

Average speed

= 80km/h

= (80÷60) km/min

= 1⅓ km/min

1⅓= 28 ÷(t -24)

since duration for return journey is from t=24 mins to t mins.

$\frac{4}{3}$ (t -24)= 28

$\frac{4}{3}$ t - 32= 28

$\frac{4}{3}$ t= 32 +28

$\frac{4}{3}$ t= 60

t= $6 0\div \frac{4}{3}$

t= 45

*Here, I assume that this is a displacement- time graph, so the distance shown is the distance of the bus from the starting point because technically if it is a distance-time graph, the distance would still increase as the bus travels the 'return journey'.

Thus, distance is decreasing after t=24 and reaches zero at time= t mins so that is the return journey. (because when the bus returns back to starting point, displacement/ distance from starting point= 0km)

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4 years ago

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zhuklara [117]

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3 years ago