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sweet-ann [11.9K]
3 years ago
6

Can someone help with this plz?

Mathematics
2 answers:
Korvikt [17]3 years ago
7 0

Answer:

Hey No worries. Its very easy.

Step-by-step explanation:

Look,

Volume= 84 in.cube

Length=6 in.

Breadth=2 in.

therefore, Height= Volume/(Length*Breadth)

=84/(6*2)

=84/12

=7

irakobra [83]3 years ago
3 0

Answer:

h = 7 in

Step-by-step explanation:

The volume (V) of a cuboid is calculated using the formula

V = lbh ( l is length, b is breadth and h is height )

Given V = 84 in³, then

lbh = 84 ← substitute l = 6 and b = 2

6 × 2 × h = 84

12h = 84 ( divide both sides by 12 )

h = 7

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Stella [2.4K]

Answer:

∠ A = 50°

Step-by-step explanation:

In a parallelogram, consecutive angles are supplementary, sum to 180°

∠ A + 130° = 180° ( subtract 130° from both sides )

∠ A = 50°

3 0
3 years ago
If you borrowed $646 at an interest rate of 5% for 6 years, how much interest will you pay? Round to the nearest tenth if necess
Snezhnost [94]

Answer:

P=646

R=5%

T=2yrs

I=?

I=PRT/100

I= 646*5*2/100

I= 64.6

Step-by-step explanation:

8 0
3 years ago
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2
Savatey [412]

Answer:

sorry

Step-by-step explanation:

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Can you help me with this question
GenaCL600 [577]

Answer:

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Step-by-step explanation:

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4 0
2 years ago
Given a focus of (4, 5) and directrix of y= -3 , find the equation of the parabola.
andrey2020 [161]
Check the picture below.

notice, the focus point is at 4,5 whilst the directrix line is at y = -3, below the focus point, meaning the parabola is vertical and opening upwards.

keeping in mind that the vertex is "p" distance from either of these fellows, then the vertex is half-way between both of them, notice in the picture, the distance from y = 5 to y = -3 is 8 units, half that is 4 units, thus the vertex 4 units from the focus or 4 units from the directrix, that puts it at (4,1), whilst "p" is 4, since the parabola is opening upwards, is a positive 4 then.

\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}
h=4\\
k=1\\
p=4
\end{cases}\implies (x-4)^2=4(4)(y-1)\implies (x-4)^2=16(y-1)
\\\\\\
\cfrac{1}{16}(x-4)^2=y-1\implies \cfrac{1}{16}(x-4)^2+1=y

8 0
3 years ago
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